In this page, integration by substitution is explained with examples.

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Consider the integration `int 2x sin x^2 dx`

We notice that `d/(dx) x^2 = 2x` is part of the integration.

Which of the following can be used to work out the integration?

- take `t=x^2` and substitute in the integrand
- take `t=x^2` and substitute in the integrand
- noticing `d/(dx) x^2 = 2x` does not help

The answer is "take `t=x^2` and substitute in the integrand".

integration `int 2x sin x^2 dx`

Substitute `t=x^2` and `dt = 2xdx`.

`int 2x sin x^2 dx`

`= int sin t dt`

`= -cos t +c` *substitute `t=x^2`*

`= -cos x^2 + c`

Considering the integration `int f(x)dx` :

If it is observed that `f(x)` can be rewritten as

`f(x) = h[g(x)] g′(x)`

then,

`int f(x)dx = int h(y)dy`

where `y=g(x)`

and `dy = g′(x)dx`

This method is "integration by substitution".

Integration by substitution : `int h(g(x)) g′(x) dx = [int h(y)dy]|_(y=g(x))`

**Integration by Substitution: ** When `f(x) = h(g(x)) g′(x)`

then,

`int f(x)dx = int h(y)dy`

where `y=g(x)`

and `dy = g′(x)dx` and so `int f(x)dx ``= [int h(y)dy]|_(y=g(x)) + c`

Given a function in parametric form `y=rcos t` and `x=rsin t`, How to proceed with the integration `int ydx`?

- Substitute `y=r cos t` and `dx = d (r cos t)`. Then, proceed with variable of integration as `t`
- Substitute `y=r cos t` and `dx = d (r cos t)`. Then, proceed with variable of integration as `t`
- the function in parametric form cannot be integrated

The answer is 'Substitute `y=r cos t` and `dx = d (r cos t)`'

Given a function in parametric form `y=rcos t` and `x=rsin t`.

`int ydx`*substituting `y=rcos t` and `dx = d(r sin t)`. the `d()` means derivative of *

`= int r cost d(r sin t)`

`= int r cos t r cos t dt`

`= int r^2 cos^2 t dt`

This can be integrated.

*Solved Exercise Problem: *

Integrate `int tan x dx`

- can be integrated with `tan x = - (cos x)′//cos x`
- can be integrated with `tan x = - (cos x)′//cos x`
- no known method to integrate

The answer is "can be integrated"

`int tanx dx`

`=int (- (cos x)′)/(cos x)`

`=-log|cos x| + c`

*Solved Exercise Problem: *

Integrate `int cot x dx`

- can be integrated with `cot x = (sin x)′//sin x`
- can be integrated with `cot x = (sin x)′//sin x`
- no known method to integrate

The answer is "can be integrated"

`int tanx dx`

`=((sin x)′)/(sin x)`

`=log|sin x| + c`

*Solved Exercise Problem: *

Integrate `int sec x dx`

- can be integrated with `sec x ``= (sec x + tan x)′``//(sec x + tan x)`
- can be integrated with `sec x ``= (sec x + tan x)′``//(sec x + tan x)`
- no known method to integrate

The answer is "can be integrated"

`int sec x dx`

`=int ((sec x + tan x)′)/(sec x + tan x) dx`

`=log|sec x + tan x|+c `

*Solved Exercise Problem: *

Integrate `int csc x dx`

- can be integrated with `csc x ``= -(csc x + cot x)′``//(csc x + cot x)`
- can be integrated with `csc x ``= -(csc x + cot x)′``//(csc x + cot x)`
- no known method to integrate

The answer is "can be integrated"

`int csc x dx`

`=int (-(csc x + cot x)′)/(csc x + cot x) dx`

`=-log|cscx+ cotx| + c`

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