Definite Integrals

Using the formula:

⇒ 3 × 2 – 2 × 2

⇒ 6 – 4 = 2

Using the formula:

⇒ log 3 + 7 – log –2+7

⇒ log |10|– log |5|

Using the formula:

Using the formula:

Let

⇒ On differentiation, we get

2 x dx = dt

⇒ Hence the question will become:

Using the formula:

⇒

⇒

⇒

⇒

⇒

Now, Using the formula:

Using the formula:

⇒

⇒ [tan^–1 (1) – tan^–1 (– 1) ]

⇒

⇒

Using the formula:

Using the formula:

(∵ log 1 = 0 )

⇒

Using the formula:

and

Using the formula:

Using the formula:

(∵ log 1 = 0 )

Using the formula :

⇒

Using the formula:

(∵ 1 – sin² x = cos² x)

Now, we know,

And,

(∵ 1 – sin² x = cos² x)

Now, we know,

And,

(∵ sec (– θ) = sec θ)

Let I =

⇒

Using the formula:

⇒

⇒

⇒

Let sin x = t. Hence, cos x dx = dt, for the second expression.

⇒

⇒

Put back t = sin(x)

We know,

⇒

⇒

⇒

Let sin x = t. Hence, cos x dx = dt. For second expression,

Put t = sin(x)

⇒

(equation 2)

From equation 2 put value of in equation 1.

⇒

⇒

⇒

⇒

First let us find,

Let sin x = t. Hence, cos x dx = dt. For second expression,

Put t = sin(x)

⇒

We know, tan x × cot x = 1

We know, tan²x = sec² x 1 1 and cot² x = cosec² x – 1

⇒

⇒

⇒

We know integration of sec²x is tanx and of cosec² x is –cotx. Therefore,

(∵ 1 + cos 2θ = 2 cos² θ)

⇒

Let 1 – sin x = t². Hence, – cos x dx = 2 t dt and cos x dx = – 2 t dt.

Let 1 – cos x = t² hence sin x dx = 2 t dt

We are asked to calculate

For this we have to apply integration by parts

Let u and v be two functions then

To choose the first function u we use “ILATE” rule

That is

I=inverse trigonometric function

L=logarithmic function

A=algebraic function

T=trigonometric functions

E=exponential function

So in this preference, the first function is choosen to make the integration simpler.

Now, In the given question x is an algebraic function and it is chosen as u (A comes first in “ILATE” rule)

So first let us integrate the equation and then let us substitute the limits in it

Therefore, now substitute the limits given:

Note that and

First we have to substitute the upper limit and then subtract the second limit value from it
)

Note that sin0= 0 and cos0=1

=0+1+0–0

=1

We are asked to calculate

For this we have to apply integration by parts

Let u and v be two functions then

To choose the first function u we use “ILATE” rule

That is

I=inverse trigonometric function

L=logarithmic function

A=algebraic function

T=trigonometric functions

E=exponential function

So in this preference, the first function is chosen to make the integration simpler.

Now, In the given question x is an algebraic function and it is chosen as u(A comes first in “ILATE” rule)

So first let us integrate the equation and then let us substitute the limits in it

Therefore, now substitute the limits given:

Note that and

First we have to substitute the upper limit and then subtract the second limit value from it

=

For this, we have to apply integration by parts

Let u and v be two functions then

To choose the first function u we use “ILATE” rule

That is

I=inverse trigonometric function

L=logarithmic function

A=algebraic function

T=trigonometric functions

E=exponential function

So in this preference,, the first function is chosen to make the integration simpler.

Now, In the given question x² is an algebraic function and it is chosen as u(A comes first in “ILATE” rule)

So first let us integrate the equation and then let us substitute the limits in it.

So now we have to substitute the limits in this equation.

And should subtract upper limit value from lower limit value

Sin , cos , sin 0 =0, cos0 = 1.

.

For this, we have to apply integration by parts

Let u and v be two functions then

To choose the first function u we use “ILATE” rule

That is

I=inverse trigonometric function

L=logarithmic function

A=algebraic function

T=trigonometric functions

E=exponential function

So in this preference,, the first function is chosen to make the integration simpler.

Now, In the given question x² is an algebraic function and it is chosen as u(A comes first in “ILATE” rule)

So first let us integrate the equation and then let us substitute the limits in it.

So now we have to substitute the limits in this equation.

And should subtract upper limit value from lower limit value

Sin, cos, sin 0 =0, cos0 = 1

For this we have to apply integration by parts

Let u and v be two functions then

To choose the first function u we use “ILATE” rule

That is

I=inverse trigonometric function

L=logarithmic function

A=algebraic function

T=trigonometric functions

E=exponential function

So in this preference, the first function is chosen to make the integration simpler.

Now, In the given question x² is an algebraic function and it is chosen as u(A comes first in “ILATE” rule)

So first let us integrate the equation and then let us substitute the limits in it.

Note that and

For this we have to apply integration by parts

Let u and v be two functions then

To choose the first function u we use “ILATE” rule

That is

I=inverse trigonometric function

L=logarithmic function

A=algebraic function

T=trigonometric functions

E=exponential function

So in this preference, the first function is chosen to make the integration simpler.

Now, In the given question x² is an algebraic function and it is chosen as u(A comes first in “ILATE” rule)

So first let us integrate the equation and then let us substitute the limits in it.

Let us recall a formula cos2x=2–1

Now substitute it

Now let us recall other formula i.e=

and

Using them we can write the equation as

On substituting these values we get

For this we have to apply integration by parts

Let u and v be two functions then

To choose the first function u we use “ILATE” rule

That is

I=inverse trigonometric function

L=logarithmic function

A=algebraic function

T=trigonometric functions

E=exponential function

So in this preference, the first function is chosen to make the integration simpler.

Now, In the given question 1 is an algebraic function and it is chosen as u(A comes first in “ILATE” rule)

So first let us integrate the equation and then let us substitute the limits in it.

Let us recall that derivative of logx is 1/x

=xlogx–x

Now let us substitute the limits

= 2 log2–2–[1log1–1]

=2log2–1

For this we have to apply integration by parts

Let u and v be two functions then

To choose the first function u we use “ILATE” rule

That is

I=inverse trigonometric function

L=logarithmic function

A=algebraic function

T=trigonometric functions

E=exponential function

So in this preference, the first function is chosen to make the integration simpler.

Now we will substitute the limits

let us assume that the given equation is L

For this we have to apply integration by parts

Let u and v be two functions then

To choose the first function u we use “ILATE” rule

That is

I=inverse trigonometric function

L=logarithmic function

A=algebraic function

T=trigonometric functions

E=exponential function

So in this preference, the first function is chosen to make the integration simpler.

We know that loge=1

Here in this question by observation we can notice that the derivative of logx is 1/x and the function integral is like

Here to solve these kinds of question let us assume logx=t

Now

Now let us change the limits

x=1 then t=0

x=e then t=1

For this we have to apply integration by parts

Let u and v be two functions then

To choose the first function u we use “ILATE” rule

That is

I=inverse trigonometric function

L=logarithmic function

A=algebraic function

T=trigonometric functions

E=exponential function

So in this preference, the first function is chosen to make the integration simpler.

Now let us substitute in the given question equation

Here we are solving the equation, recall is the derivative of log(x+2) and splitting the second one

Note that log4=2log2 and log1=0

If the equation is in this form then convert the numerator as sum of derivative of denominator and some constant

Here we know that denominator derivative is 10x

So to get it in the numerator multiply and divide by 5

Now you get the equation as

We already know that derivative of logx is 1/x

Using that here derivative of

And derivative of

So

Now substitute limits 1 and 0

Since it is a quadratic equation we are trying to make it a complete square

=here the equation is in the form of integral of the integral is equal to

Here let us assume that t=x–

So that dx=dt

When x=0 t=–1/2

And when x=–1/2, t=3/2

]

]

)]

Now rationalize the denominator

We get as

Since the denominator is a quadratic equation let us make it in form of a perfect square

Now the equation

derivative of using this we can write it as

}

To solve these kinds of equations we generally take x=

So now here let

So now

Now change the limits

X=0 then =0

X=1 then =

So it is equal to

now use formula sin2x=2sinxcosx

Now use formula that

)

Now let us recall other formula :

and

Now recall that sin0=0, cos0=1

Here the equation is of form that a quadratic equation is in the root so now to solve this make the equation in the root in the form of a²–x²,a²+x²

Here

=4–(x–1)²

=2²–(x–1)²

Now just recall a formula that is derivative of

Here a=2 and x =x–1

Now we get

Here first we are converting the quadratic equation in to a perfect square

)

Now just recall a formula that is derivative of

Here a=2 and x =x–2

Now we get

)

Now denominator is in a quadratic form so let us make it in other form

Recall a formula

So now .

To solve this let us assume that 2x+1=

2 dx=2t dt

So now x=1,

X=4, t=3

So now after substitution we get

Now let us recall other formula i.e

Now to make it simpler problem let us expand using binomial theorem

So

Let us also recall other formula i.e.,

So now

For this we have to apply integration by parts

Let u and v be two functions then

To choose the first function u we use “ILATE” rule

That is

I=inverse trigonometric function

L=logarithmic function

A=algebraic function

T=trigonometric functions

E=exponential function

So in this preference, the first function is chosen to make the integration simpler.

Now, In the given question x² is an algebraic function and it is chosen as u(A comes first in “ILATE” rule)

So first let us integrate the equation and then let us substitute the limits in it.

Here we are expanding only first integral first

First split the integral

Now integrate by parts the first one

For this we have to apply integration by parts

Let u and v be two functions then

To choose the first function u we use “ILATE” rule

That is

I=inverse trigonometric function

L=logarithmic function

A=algebraic function

T=trigonometric functions

E=exponential function

So in this preference, the first function is chosen to make the integration simpler.

Now, In the given question x² is an algebraic function and it is chosen as u(A comes first in “ILATE” rule)

So first let us integrate the equation and then let us substitute the limits in it.

Remember and

First split the integral

Now integrate by parts the first one

For this we have to apply integration by parts

Let u and v be two functions then

To choose the first function u we use “ILATE” rule

That is

I=inverse trigonometric function

L=logarithmic function

A=algebraic function

T=trigonometric functions

E=exponential function

So in this preference, the first function is chosen to make the integration simpler.

Now, In the given question x² is an algebraic function and it is chosen as u(A comes first in “ILATE” rule)

So first let us integrate the equation and then let us substitute the limits in it.

Remember and

Now using the formula

Sin2x=2sinxcosx

Here we know that derivative of is

And it is in the form of so the equation integral will be function

We know that

So the equation will be

We know that cos45=sin45=

Substitute it

Remember that and

Now integrate by parts
For this we have to apply integration by parts

Let u and v be two functions then

To choose the first function u we use “ILATE” rule

That is

I=inverse trigonometric function

L=logarithmic function

A=algebraic function

T=trigonometric functions

E=exponential function

So in this preference, the first function is chosen to make the integration simpler.

Now, In the given question x² is an algebraic function and it is chosen as u(A comes first in “ILATE” rule)

So first let us integrate the equation and then let us substitute the limits in it.

}

=0–0=0

Now let us use integration by parts

For this we have to apply integration by parts

Let u and v be two functions then

To choose the first function u we use “ILATE” rule

That is

I=inverse trigonometric function

L=logarithmic function

A=algebraic function

T=trigonometric functions

E=exponential function

So in this preference, the first function is chosen to make the integration simpler.

Now, In the given question x² is an algebraic function and it is chosen as u(A comes first in “ILATE” rule)

So first let us integrate the equation and then let us substitute the limits in it.

Remember that and

limit 0 to pi

with limits 0 to pi

}

Now let it be taken as I

dx

Now rationalize the denominator

Let us also recall formula

Remember derivative of

So using that

Substitute upper limit and then subtract the lower limit from it

=–(log3–log2)+2(log4–log3)

=–3log3+5log2

Let

Remember that and

Because we have a formula

Remember that and

]

=0

Let 2x=t then 2dx=dt

When x=1 t=2

And when x=2,t=4

We can observe here that

Derivative of

Now it is in the form

So the integral will be

Let us solve the denominator

(

=3x–x²–2

Now just recall a formula that is derivative of

Given that ,k=?

derivative of

2k=1

Let us also recall formula

a=2

hence a=2.

We know that

Now substitute that in the equation

We get

We already know that integral of sinx is –cosx

Let

Let us recall that

And

Recall: and

=4(0+1+1–0)

=8

We know that

Now substitute them in the equation.

Let us recall that

Again using

Here we are using reduction formula of sinx

For n=2

[limits 0,]

[limits 0,]

Now substitute limits

Now

Now let us use integration by parts

For this we have to apply integration by parts

Let u and v be two functions then

To choose the first function u we use “ILATE” rule

That is

I=inverse trigonometric function

L=logarithmic function

A=algebraic function

T=trigonometric functions

E=exponential function

So in this preference, the first function is chosen to make the integration simpler.

Now, In the given question x² is an algebraic function and it is chosen as u(A comes first in “ILATE” rule)

So first let us integrate the equation and then let us substitute the limits in it.

dx

Integral and integral of

Recall:

We know integral of cosx is sinx

Now arranging denominator, we get as

Now recall integral

And,

Given definite integral is:

Let us assume ……(1)

Assume y = x²+1

Differentiating w.r.t x on both sides we get,

d(y) = d(x² + 1)

dy = 2x dx

……(2)

The upper limit for Integral

X = 4 ⇒ y = 4² + 1

Upper limit: y = 17……(3)

The lower limit for Integral

X = 2 ⇒ y = 2² + 1

Lower limit: y = 5 ……(4)

Substituting (2),(3),(4) in the eq(1), we get,

We know that:

We know that:

[here f’(x) is derivative of f(x))

We know that:

Given Definite Integral can be assumed as:

……(1)

Let us assume y = 1 + log(x)

Differentiating w.r.t x on both sides we get

⇒ d(y) = d(1 + log( x ))

……(2)

Lower limit for Definite Integral:

⇒ x = 1 ⇒ y = 1 + log 1

⇒ y = 1 ……(3)

Upper limit for Definite Integral:

⇒ x = 2 ⇒ y = 1 + log2

⇒ y = 1 + log2 ……(4)

Substituting (2),(3),(4) in the eq(1) we get,

We know that:

We know that:

[here f’(x) is derivative of f(x))

We know that loge=1 and loga+logb=logab

Given Definite Integral can be written as:

……(1)

Let us assume y = 9x²–1

Differentiating w.r.t x on both sides we get

⇒ d(y) = d(9x²–1)

⇒ dy = 18 x dx

……(2)

Upper limit for Definite Integral:

⇒ x = 1 ⇒ y = (9 × 1²) – 1

⇒ y = 8……(3)

Lower limit for Definite Integral:

⇒ x = 2 ⇒ y = (9 × 2²) – 1

⇒ y = 35……(4)

Substituting (2),(3),(4) in the eq(1), we get,

We know that:

We know that:

[here f’(x) is derivative of f(x))

We know that:

Given Definite Integral can be written as:

……(1)

We know that:

And

Let us find the value of 5cosx+3sinx

We know that: 1+tan²x = sec²x

……(2)

Substituting (2) in (1) we get,

Let us assume:

Differentiating on both sides w.r.t x we get,

……(3)

The upper limit for the Definite Integral:

⇒ t=1……(4)

The lower limit for the Definite Integral:

⇒ t=0……(5)

Substituting (3),(4),(5) in the eq(1) we get,

We need to convert the denominator into standard forms

We know that:

In this problem the values,

Using these values and the standard result, we get,

We know that:

[here f’(x) is derivative of f(x))

We know that:

Given Definite integral can be written as:

(1)

Let us assume y = a²+x²

Differentiating w.r.t x on both sides we get,

⇒ d(y) = d(a²+x²)

⇒ dy = 2xdx

……(2)

Upper limit for the Definite Integral:

⇒ x=a ⇒ y = a²+a²

⇒ y=2a²……(3)

Lower limit for the Definite Integral:

⇒ x=0 ⇒ y = a²+0²

⇒ y = a²……(4)

Substituting (2),(3),(4) in the eq(1), we get,

We know that:

We know that:

[here f’(x) is derivative of f(x))

⇒ I(x) = (2a² )^1/2 – (a² )^1/2

⇒ I(x) = √2 a – a

⇒ I(x) = a(√2–1)

Given Definite Integral can be written as:

......(1)

Let us assume y = e^x

Differentiating w.r.t x on both sides we get,

⇒ d(y) = d(e^x)

⇒ dy = e^xdx ......(2)

Upper limit for the Definite Integral:

⇒ x = 1 ⇒ y = e¹

⇒ y = e(3)

Lower limit for the Definite Integral:

⇒ x = 0 ⇒ y = e⁰

⇒ y = 1(4)

Substituting (2),(3),(4) in the eq(1) we get,

We know that:

We know that:

[here f’(x) is derivative of f(x))

Given Definite Integral can be written as:

(1)

Let us assume y = x²

Differentiating w.r.t x on both sides we get,

⇒ d(y) = d(x²)

⇒ dy = 2xdx

……(2)

Upper limit for the Definite Integral:

⇒ x = 1 ⇒ y = 1²

⇒ y = 1 ……(3)

Lower limit for the Definite Integral:

⇒ x = 0 ⇒ y = 0²

⇒ y = 0 ……(4)

Substituting (2),(3),(4) in the eq(1), we get,

We know that: ∫ e^xdx = e^x+c

We know that:

[here f’(x) is derivative of f(x))

Given Definite Integral can be written as:

……(1)

Let us assume y = logx

Differentiating w.r.t x on both sides

⇒ d(y) = d(logx)

……(2)

Upper limit for the Definite Integral:

⇒ x = 3 ⇒ y = log(3)

⇒ y = log3……(3)

Lower limit for the Definite Integral:

⇒ x = 1 ⇒ y = log(1)

⇒ y = 0……(4)

Substituting (2),(3),(4) in the eq(1) we get,

We know that ∫ cos x dx = sin x + c

We know that:

here f’(x) is derivative of f(x))

⇒ I(x) = sin(log3) – sin(0)

⇒ I(x) = sin(log3) – 0

⇒ I(x) = sin(log3)

Given Definite Integral can be written as:

……(1)

Let us assume y = x²

Differentiating w.r.t x on both sides we get,

⇒ d(y) = d(x²)

⇒ dy = 2xdx……(2)

Lower limit for the Definite Integral:

⇒ x = 0 ⇒ y = 0²

⇒ y = 0……(3)

Upper limit for the Definite Integral:

⇒ x = 1 ⇒ y = 1²

⇒ y = 1……(4)

Substituting (2),(3),(4) in the eq(1), we get,

We know that:

We know that

[here f’(x) is derivative of f(x))

⇒ I(x) = tan^-1(1) – tan^-1(0)

Given Definite Integral can be written as:

…… (1)

Let us assume x = a sinθ

Differentiating w.r.t x on both sides we get,

⇒ d(x) = d(a sin θ)

⇒ dx = a cos θ dθ ……(2)

Let us find the value of

(∵ 1 – sin²θ = cos²θ)

……(3)

Lower limit for the Definite Integral:

⇒ θ = sin^-1(0)

⇒ θ = 0……(4)

Upper limit for the Definite Integral:

⇒ θ = sin^-1(1)

……(5)

Substituting (2),(3),(4),(5) in eq(1) we get,

We know that cos2θ = 2cos²θ – 1

Then

Using these result for the integration, we get,

We know that:

∫ adx = ax + c and also

We know that:

[here f’(x) is derivative of f(x)).

We know that sinnπ = 0 (n∈I)

Given Definite Integral can be written as:

Let us assume sinϕ = t,

Differentiating w.r.t ϕ on both sides we get,

⇒ d(sinϕ) = d(t)

⇒ dt = cosϕ dϕ……(2)

Upper limit for the Definite Integral:

⇒ t = 1……(3)

Lower limit for the Definite Integral:

⇒ ϕ=0 ⇒ t = sin(0)

⇒ t = 0……(4)

We know that cos²ϕ = 1-sin²ϕ

⇒ cos²ϕ = 1 – t²……(5)

Substituting (2),(3),(4),(5) in the eq(1), we get,

We know that:

We know that:

[here f’(x) is derivative of f(x))

Given Definite Integral can be written as:

Let us assume y = sinx,

Differentiating on both sides w.r.t x we get,

⇒ d(y) = d(sinx)

⇒ dy = cosxdx……(2)

Upper limit for the Definite Integral:

⇒ y = 1……(3)

Lower limit for the Definite Integral:

⇒ x = 0 ⇒ y = sin(0)

⇒ y = 0……(4)

Substituting (2),(3),(4) in the eq(1) we get,

We know that:

We know that:

[here f’(x) is derivative of f(x))

Given Definite Integral can be written as:

Let us assume 1+cosθ=y

Differentiating w.r.t θ on both sides we get,

⇒ d(1+cosθ) = d(y)

⇒ -sinθdθ = dy

⇒ sinθdθ = -dy……(2)

Upper limit for the Definite Integral

⇒ y = 1……(3)

Lower limit for the Definite Integral:

⇒ θ = 0 ⇒ y = 1+cos(0)

⇒ y = 1+1

⇒ y = 2……(4)

Substituting (2),(3),(4) in the eq(1), we get,

We know that:

We know that:

We know that:

[here f’(x) is derivative of f(x))

Given Definite Integral can be written as:

Let us assume 3+4sinx = y

Differentiating w.r.t x on both sides we get,

⇒ d(3+4sinx) = d(y)

⇒ 4cosxdx = dy

Lower limit for the Definite Integral:

⇒ x = 0 ⇒ y = 3+4sin(0)

⇒ y = 3 + 0

⇒ y = 3……(3)

Upper limit for the Definite Integral:

Substituting (2),(3),(4) in the eq(1) we get,

We know that:

We know that:

[here f’(x) is derivative of f(x))

Given Definite Integral can be written as:

Let us assume tan^-1x = y

Differentiating w.r.t x on both sides we get,

⇒ d(tan^-1x) = d(y)

Upper limit of the Definite Integral:

⇒ x = 1 ⇒ y = tan^-1(1)

Lower limit of the Definite Integral:

⇒ x = 0 ⇒ y = tan^-1(0)

⇒ y = 0…… (4)

Substituting (2),(3),(4) in the eq(1) we get,

We know that:

We know that:

[here f’(x) is derivative of f(x))

Given Definite Integral can be written as:

Let us assume x+2 = y

Then, x = y-2 ……(2)

Differentiating on both side w.r.t x we get,

⇒ d(x+2) = d(y)

⇒ dx = dy ……(3)

Upper limit for the Definite Integral:

⇒ x = 2 ⇒ y = 2+2

⇒ y = 4…… (4)

Lower limit for the Definite Integral:

⇒ x = 0 ⇒ y = 0+2

⇒ y = 2…… (5)

Substituting (2),(3),(4),(5) in the eq(1) we get,

We know that:

We know that:

[here f’(x) is derivative of f(x))

Given Definite Integral can be written as:

Let us assume x = tany

Differentiating w.r.t x on both sides we get,

⇒ d(x) = d(tany)

⇒ dx = sec²ydy……-(2)

Then

We know that:

Now,

Upper limit for the Definite Integral:

⇒ x = 1 ⇒ y = tan^-1(1)

Lower limit for the Definite Integral:

⇒ x = 0 ⇒ y = tan^-1(0)

⇒ y = 0…… (5)

Substituting (2),(3),(4),(5) in (1) we get,

We know that the By-partss integration is:

Now applying by parts Integration:

We know that: ∫ sec²xdx = tanx + C

We know that:

[here f’(x) is derivative of f(x))

We know that: ∫ tanxdx = -log(cosx) + C

We know that: log(a^b) = bloga

Given Definite Integral can be written as:

Let us assume, y = sin²x

Differentiating w.r.t x on both sides we get,

⇒ d(y) = d(sin²x)

⇒ dy = 2sinxcosxdx

Upper limit for the Definite Integral:

⇒ y = 1…… (3)

Lower limit for the Definite Integral:

⇒ x = 0 ⇒ y = sin²0

⇒ y = 0……(4)

Substituting (2),(3),(4) in the eq(1) we get,

We know that:

We know that:

[here f’(x) is derivative of f(x))

Given Definite Integral can be written as:

We know that:

Substituting these value in (1) we get,

We know that: 1+tan²x = sec²x

Let us assume,

Differentiating w.r.t x on both sides we get,

Upper limit for the Definite Integral:

Lower limit for the Definite Integral:

Substituting (2),(3),(4) in eq(1), we get,

We know that:

We know that:

[here f’(x) is derivative of f(x)).

We know that: Log = log(a) – log(b)

Given Definite can be written as:

We know that:

We know that: 1+tan²x = sec²x

Let us assume,

Differentiating w.r.t x on both sides we get,

Upper limit for the Definite Integral:

Lower limit for the Definite Integral:

Substituting (2),(3),(4) in the eq(1) we get,

We know that:

We know that:

[here f’(x) is derivative of f(x)).

We know that:

Given Definite Integral can be written as:

Let us write numerator in terms of the denominator for easy calculation,

⇒ sinx = K(sinx+cosx) + L(cosx – sinx)

⇒ sinx = sinx(K-L) +cosx(K+L)

Comparing coefficients of corresponding terms on both sides we get,

⇒ K + L =0

⇒ K – L =1

On solving these two equations we get,

L= and K = .

So numerator can be written as:

Substituting these values in(1) we get,

We know that:

We know that:

[here f’(x) is derivative of f(x)).

Given Definite Integral can be written as:

We know that:

We know that: 1+tan²x = sec²x

Let us assume,

Differentiating w.r.t x on both sides we get,

Upper limit for the Definite Integral:

⇒ y = 1+∞

⇒ y = ∞……(3)

Lower limit for the Definite Integral:

⇒ x = 0 ⇒ y = 1+tan(0)

⇒ y = 1+0

⇒ y = 1……(4)

Substituting (2),(3),(4) in the eq(1), we get,

We know that:

We know that:

[here f’(x) is derivative of f(x)).

⇒ I(x) = tan^-1(∞) – tan^-1(0)

Given Definite Integral can be written as:

We will find the value of ∫ tan^-1xdx using by parts rule

Let us find the value of ∫ tan^-1xdx

⇒ ∫ tan^-1x dx = ∫ 1.tan^-1xdx

We substitute this result in the Definite Integral:

We know that:

[here f’(x) is derivative of f(x)).

Given Definite Integral can be written as:

Let us find the value of using by parts integration,

Now we substitute this result in the Definite Integral:

We know that:

[here f’(x) is derivative of f(x)).

Given Definite Integral can be written as:

We know that:

Substituting in the Definite Integral we get,

Let us assume, y = sinx – cosx

Differentiating w.r.t x on both sides we get,

⇒ d(y) = d(sinx – cosx)

⇒ dy = (cosx + sinx)dx……(2)

Upper limit for the Definite Integral:

⇒ y = 0…… (3)

Lower limit for the Definite Integral:

⇒ x = 0 ⇒ y = sin(0) – cos(0)

⇒ y = 0 – 1

⇒ y = -1……(4)

Substituting (2),(3),(4) in the eq(1) we get,

We know that:

We know that:

[here f’(x) is derivative of f(x))

Given Definite Integral can be written as:

We know that: 1+ cos2x = 2cos²x and

Let us assume, y = tanx

Differentiating w.r.t x on both sides we get,

⇒ d(y) = d(tanx)

⇒ dy = sec²xdx……(2)

Upper limit for the Definite Integral:

⇒ y = 1……(3)

The lower limit for the Definite Integral:

⇒ y = 0……(4)

Substituting (2),(3),(4) in eq(1) we get,

We know that:

We know that:

[here f’(x) is derivative of f(x)).

Given Definite Integral can be written as:

We know that:

We know that: 1+tan²x = sec²x

Let us assume,

Differentiating w.r.t x on both the sides we get,

The upper limit for the Definite Integral:

⇒ y = ∞……(3)

Lower limit for the Definite Integral:

⇒ y = 0……(4)

Substituting (2),(3),(4) in the eq(1) we get,

We know that:

We know that:

[here f’(x) is derivative of f(x)).

Given Definite Integral can be written as:

Let us assume y = tanx

Differentiating w.r.t x on both sides we get,

⇒ d(y) = d(tanx)

⇒ dy = sec²xdx……(2)

Upper limit for the Definite Integral:

⇒ y = ∞……(3)

Lower limit for the Definite Integral:

⇒ x = 0 ⇒ y = tan(0)

⇒ y = 0……(4)

Substituting (2),(3),(4) in the eq(1) we get,

We know that:

We know that:

[here f’(x) is derivative of f(x)).

Given Definite Integral can be written as:

We know that sin 2x = 2 sinx cosx and 1 + cos2x = 2 cos²x

Applying by-parts integration for 1^st term only

We know that:

[here f’(x) is derivative of f(x)).

Given Definite Integral can be written as:

……(1)

Let us assume y = tan^-1x

Differentiating w.r.t x on both sides we get,

⇒ d(y) = d(tan^-1x)

……(2)

Upper limit for the Definite Integral:

⇒ x = 1 ⇒ y = tan^-1(1)

……(3)

Lower limit for the Definite Integral:

⇒ x = 0 ⇒ y =tan^-1(0)

⇒ y = 0……-(4)

Substitute (2),(3),(4) in the eq(1) we get,

We know that:

We know that:

[here f’(x) is derivative of f(x)).

Let

In the denominator, we have sin 2x = 2 sin x cos x

Note that we can write 2 sin x cos x = 1 – (1 – 2 sin x cos x)

We also have sin²x + cos²x = 1

⇒ 1 – 2 sin x cos x = sin²x + cos²x – 2 sin x cos x

⇒ sin 2x = 1 – (sin x – cos x)²

So, using this, we can write our integral as

Now, put sin x – cos x = t

⇒ (cos x + sin x) dx = dt (Differentiating both sides)

When x = 0, t = sin 0 – cos 0 = 0 – 1 = -1

When,

So, the new limits are -1 and 0.

Substituting this in the original integral,

Recall,

Let

We will use integration by parts.

Recall,

Here, take f(x) = tan^-1x and g(x) = x

We have,

Now,

Substituting these values, we evaluate the integral.

We can write,

Recall,

Let

In the denominator, we have x⁴ + x² + 1

Note that we can write x⁴ + x² + 1 = (x⁴ + 2x² + 1) – x²

We have x⁴ + 2x² + 1 = (1 + x²)²

⇒ x⁴ + x² + 1 = (1 + x²)² – x²

So, using this, we can write our integral as

Dividing numerator and denominator with x², we have

Put,

(Differentiating both sides)

So, the new limits are ∞ and 2.

Substituting this in the original integral,

Recall,

Let

Put 1 + x² = t

⇒ 2xdx = dt (Differentiating both sides)

When x = 0, t = 1 + 0² = 1

When x = 1, t = 1 + 1² = 2

So, the new limits are 1 and 2.

In numerator, we can write 24x³dx = 12x² × 2xdx

But, x² = t – 1 and 2xdx = dt

⇒ 24x³dx = 12(t – 1)dt

Substituting this in the original integral,

Recall

Let

Put x – 4 = t³

⇒ dx = 3t²dt (Differentiating both sides)

When x = 4, t³ = 4 – 4 = 0 ⇒ t = 0

When x = 12, t³ = 12 – 4 = 8 ⇒ t = 2

So, the new limits are 0 and 2.

We can write x = t³ + 4

Substituting this in the original integral,

Recall,

Let

We will use integration by parts.

Recall,

Here, take f(x) = x² and g(x) = sin x

Now,

⇒ f’(x) = 2x

Substituting these values, we evaluate the integral.

Let

We use integration by parts again.

Here, take f(x) = x and g(x) = cos x

Now,

⇒ f’(x) = 1

Using these values in equation for I₁

Substituting I₁ in I, we get

⇒ I = (0 + π + 0) – (2)

⇒ I = π – 2

Let

As we have the trigonometric identity, to evaluate this integral we use x = cos 2θ

⇒ dx = –2sin(2θ)dθ (Differentiating both sides)

When x = 0, cos 2θ = 0 ⇒ 2θ = ⇒ θ =

When x = 1, cos 2θ = 1 ⇒ 2θ = 0 ⇒ θ = 0

So, the new limits are and 0.

Substituting this in the original integral,

We have and sin 2θ = 2 sin θ cos θ

But,

Let

As we have the trigonometric identity 1 + tan²θ = sec²θ, to evaluate this integral we use x = tan θ

⇒ dx = sec²θ dθ (Differentiating both sides)

When x = 0, tan θ = 0 ⇒ θ = 0

So, the new limits are 0 and.

Substituting this in the original integral,

We have cos²θ – sin²θ = cos 2θ

Let

Put x⁵ + 1 = t

⇒ 5x⁴dx = dt (Differentiating both sides)

When x = –1, t = (–1)⁵ + 1 = 0

When x = 1, t = 1⁵ + 1 = 2

So, the new limits are 0 and 2.

Substituting this in the original integral,

Recall

Let

Dividing numerator and denominator with cos²x, we have

Put tan x = t

⇒ sec²x dx = dt (Differentiating both sides)

When x = 0, t = tan 0 = 0

When x = , t = tan = ∞

So, the new limits are 0 and ∞.

Substituting this in the original integral,

Multiplying numerator and denominator with 3, we have

Now, we can write

Substituting this in the original integral,

Recall

Let

Put sin 2t = x

⇒ 2cos(2t)dt = dx (Differentiating both sides)

When t = 0, x = sin 0 = 0

So, the new limits are 0 and 1.

Substituting this in the original integral,

Recall

Let

Put 5 – 4 cos θ = x

⇒ 4sin(θ)dθ = dx (Differentiating both sides)

When θ = 0, x = 5 – 4 cos 0 = 5 – 4 = 1

When θ = π, x = 5 – 4 cos π = 5 – (–4) = 9

So, the new limits are 1 and 9.

Substituting this in the original integral,

Recall

Let

Put cos 2θ = x

⇒ –2sin(2θ)dθ = dx (Differentiating both sides)

When θ = 0, x = cos 0 = 1

So, the new limits are 1 and.

Substituting this in the original integral,

Recall

Let

Put

(Differentiating both sides)

So, the new limits are 0 and π.

Substituting this in the original integral,

But,

Let

Put 1 + log x = t

When x = 1, t = 1 + log 1 = 1

When x = 2, t = 1 + log 2

So, the new limits are 1 and 1 + log 2.

Substituting this in the original integral,

Recall

Let

Note that we can write cos⁵x = cos⁴x × cos x

⇒ cos⁵x = (cos²x)² × cos x

We also have sin²x + cos²x = 1

⇒ cos⁵x = (1 – sin²x)²cos x

So,

Put sin x = t

⇒ cos x dx = dt (Differentiating both sides)

When x = 0, t = sin 0 = 0

So, the new limits are 0 and 1.

Substituting this in the original integral,

Recall

Let