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8. Integration by Trigonometric Substitution

by M. Bourne

In this section, we see how to integrate expressions like


Depending on the function we need to integrate, we substitute one of the following trigonometric expressions to simplify the integration:

For `sqrt(a^2-x^2)`, use ` x =a sin theta`

For `sqrt(a^2+x^2)`, use ` x=a tan theta`

For `sqrt(x^2-a^2)`, use `x=a sec theta`

After we use these substitutions we'll get an integral that is "do-able".

Take note that we are not integrating trigonometric expressions (like we did earlier in Integration: The Basic Trigonometric Forms and Integrating Other Trigonometric Forms and Integrating Inverse Trigonometric Forms.

Rather, on this page, we substitute a sine, tangent or secant expression in order to make an integral possible.

Example 1



We can write the question as `int(dx)/((3^2+x^2)^(3//2))`

It's now in the form of the second substitution suggestion given above, that is:

For `sqrt(a^2+x^2)`, use `x=a tan theta`,

with `a=3`.

So we'll put `x=3 tan theta` and this gives `dx=3 sec^2 theta d theta`

We make the first substitution and simplify the denominator of the question before proceeding to integrate.

We'll need to use the following:

`(a^2)^(3//2) = a^3`.

Here's a number example demonstrating this expression:

`9^(3//2) = (sqrt9)^3 = 3^3 = 27`

This is a well-known trigonometric identity:

`tan^2 θ + 1 = sec^2 θ`

So we have:

`(x^2+9)^(3//2)=((3 tan theta)^2+9)^(3//2)`

`=(9 tan^2 theta+9)^(3//2)`

`=(9[tan^2 theta+1])^(3//2)`

`=9^(3//2)[tan^2 theta+1]^(3//2)`

`=27[sec^2 theta]^(3//2)`

`=27[sec theta]^3`

`=27 sec^3 theta`

Now, substituting

`dx=3 sec^2 theta d theta`


`(x^2+9)^(3//2)=27 sec^3 theta`

into the given integral gives us:

`int(dx)/((x^2+9)^(3//2))=int(3 sec^2 theta d theta)/(27 sec^3 theta)`

`=1/9int(d theta)/(sec theta)`

`=1/9int cos theta d theta`

`=1/9 sin theta+K`

We now need to get our answer in terms of x (since the question was in terms of x).

Since we let `x = 3 tan θ`, we get

`tan theta=x/3`

and we can draw a triangle to find the expression for `sin θ` in terms of `x`:`

From the triangle we see that

`sin theta=x/sqrt(x^2+9)`

Therefore, we can conclude that the answer for our integral is `1/9` times this last expression.:

`int(dx)/((x^2+9)^(3//2))=1/9sin theta+K`



Example 2



`int_4^5(sqrt(x^2-16))/x^2 dx`

This question contains a square root which is in the form of the 3rd substitution suggestion given at the top, that is:

For `sqrt(x^2-a^2)`, use `x=a sec theta`

So we have `a=4` and we let

`x = 4 sec θ`

and this gives

`x^2= 16 sec^2 θ`


`dx = 4 sec θ tan θ d θ`

Simplifying the square root part:

`sqrt(x^2-16) =sqrt(16 sec^2 theta-16)`



`=4 tan theta`


`dx = 4 sec theta tan theta d theta,`

`x^2= 16 sec^2 theta` and

`sqrt(x^2-16)=4 tan theta`

into the given integral gives us the following. (We take the indefinite case first and then do the substitution of upper and lower limits later, to make the writing a bit easier.)

`int(sqrt(x^2-16))/x^2dx =int((4 tan theta))/(16 sec^2 theta)(4 sec theta tan theta) d theta`

`=int(16 tan^2 theta sec theta)/(16 sec^2 theta) d theta`

`=int(tan^2 theta)/(sec theta) d theta`

`=int(sec^2theta-1)/(sec theta) d theta`

`=int((sec^2 theta)/(sec theta)-1/(sec theta)) d theta`

`=int (sec theta-cos theta) d theta`

`=[ln |sec theta+tan theta|-sin theta]+K`

Now, our question was a definite integral, so we need to either re-express our answer in terms of the original variable , `x`, or we could work it using `theta`.

Changing back to `x`

Earlier, we let `x = 4 sec θ`, so we get `sec theta=x/4` (or `cos theta = 4/x`).

We draw an appropriate triangle like we did earlier:

x 4 θ

Triangle to find `sin theta` and `tan theta` in terms of `x`.

We can see that:

`tan theta=(sqrt(x^2-16))/4`


`sin theta=(sqrt(x^2-16))/x`

Therefore, we can conclude that:

`int_4^5(sqrt(x^2-16))/x^2 dx =[ln|sec theta+tan theta|-sin theta]_(theta=?)^(theta=?)`

`=[ln|x/4+(sqrt(x^2-16))/4|` `{:-(sqrt(x^2-16))/x]_4^5`




(I've put `theta=?` as the upper and lower limits in the first line above because we don't know those limits in terms of `theta`, and we don't need to calculate them since we revert to `x` as our variable.)

Leaving it in terms of `theta`

Since `sec theta=x/4`, then as `x` ranges from `4` to `5`, then `sec theta` will range from `1` to `1.25`.

So the required upper and lower limits for `theta` (these are the missing question mark "`theta=?`" values in the above answer) will be

`theta="arcsec"(1)= 0`



Returning to our answer in `theta`, and substituting our upper and lower values gives:

`[ln |sec theta+tan theta|-sin theta]_(theta=0)^(theta=0.6435011)`

`=[ln |sec 0.6435011+tan 0.6435011|` `{:-sin 0.6435011]` `-[ln |sec 0+tan 0|-sin 0]`

`=0.09315`, which is the same as our earlier answer.

In this example, both approaches (leaving it in terms of `theta` or changing back to `x`) is about the same amount of work.


Integrate each of the given functions:

1. `intsqrt(16-x^2)dx`


`intsqrt(16-x^2) dx`

This question is in the form of the first substitution suggestion in this section, that is,

For `sqrt(a^2-x^2)`, use ` x =a sin theta`

So we have `a=4`, `x= 4 sin θ`, and `dx = 4 cos θ dθ`.

Substituting and simplifying the square root part first:

`sqrt(16-x^2) =sqrt(16-16 sin^2 theta)`

`=sqrt(16(1-sin^2 theta))`

`=4sqrt(cos^2 theta)`

`=4 cos theta`

Substituting into the integral gives:

`intsqrt(16-x^2) dx =int4 cos theta(4 cos theta d theta) `

` =int16 cos^2 theta d theta`

`=16int1/2(cos 2 theta+1)d theta`

`=8int(cos 2 theta+1)d theta`

`=8(1/2sin 2 theta+theta)+K`

`=8(sin theta cos theta + theta)+K`

`=8(x/4(sqrt(16-x^2))/4+arcsin {:x/4:})+K`

`=(xsqrt(16-x^2))/2+8 arcsin{:x/4:}+K`

The second-last step comes from drawing a triangle, using `sin theta = x/4` in this case, as follows:

4 x θ

Triangle to find `theta`, `sin theta` and `cos theta` in terms of `x`.

Quite often we can get different forms of the same final answer! That is, math software (or another human) can produce an answer which is actually correct, but in a different formto the one given here since. If this happens, don't panic! Just check your solution perhaps by substituting various values for `x`, or (better), drawing the graph using software.

2. `int(3 dx)/(xsqrt(4-x^2))`


`int(3 dx)/(xsqrt(4-x^2))`

This contains a `sqrt(a^2-x^2)` term, so we will use a substitution of `x =a sin theta`.

So `a=2`, and we let `x = 2 sin θ`, so `dx = 2 cos θ dθ`.

Substituting and simplifying the square root gives:

`sqrt(4-x^2) =sqrt(4-4 sin^2 theta)`

`=sqrt(4(1-sin^2 theta))`

`=2sqrt(cos^2 theta)`

`=2 cos theta`

This time our triangle will use `sin theta = x/2`, as follows:

2 x θ

Triangle to find `csc theta` and `cot theta` in terms of `x`.

Substituting everything into the integral gives:

`int(3 dx)/(xsqrt(4-x^2)) = int(3(2 cos theta\ d theta))/((2 sin theta)(2 cos theta))`

`=3/2int(d theta)/(sin theta)`

`=3/2intcsc theta d theta`

`=3/2ln |csc theta-cot theta|+K`

`=3/2ln |2/x-(sqrt(4-x^2))/x|+K`


3. `int(dx)/(sqrt(x^2+2x))`



Firstly, note that


If we put `u = x + 1`, then `du = dx` and our integral becomes:


Now, we use `u = sec θ` and so `du = sec θ tan θ dθ`

The square root becomes:

`sqrt(u^2-1)` `=sqrt(sec^2theta-1)` `=sqrt(tan^2 theta)` `=tan theta`

The triangle in this case starts with `x+1=sec theta` (that is, `cos theta = 1/(x+1)`), and is as follows:

x + 1 1 θ

Triangle to find `sec theta` and `tan theta` in terms of `x`.

Returning to our integral, we have:

`int(dx)/(sqrt(x^2+2x)) =int(du)/(sqrt(u^2-1))`

`=int(sec theta tan theta d theta)/(tan theta)`

`=int sec theta d theta`

`=ln |sec theta+tan theta|+K`

`=ln |x+1+sqrt(x^2+2x)|+K`

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