# Integration techniques | Trigonometric substitution

When none of the techniques you’ve learnt so far (u-substitution 1, 2,3,4 and integration by parts seem to work, you might consider trigonometric substitution. Basically, if you have a function that reminds you of a trigonometric identity, you let that function be equal to the result of the identity, for example cos²x from 1-sin²x, and then differentiate. To make things more clear I’m going to use this example:

$\int\sqrt{1-x^2}\,&space;dx$

U substitution and integration by parts are not going to be very helpful. What we are gonna do instead is let x=sin(theta) so that 1-sin²(theta)=cos²(theta).

$x=\sin(\theta)\;&space;\;&space;\;&space;\;&space;dx=\cos(\theta)&space;\,&space;d\theta$ $\int&space;\sqrt{1-\sin^2(\theta)}\,&space;\cos(\theta)d\theta=\int&space;\sqrt{\cos^2(\theta)}\,&space;\cos(\theta)d\theta=\int&space;\cos^2(\theta)d\theta$

Now we can integrate by parts:

$u=\cos(\theta)\;&space;\;&space;\;&space;\;&space;\;&space;dv=\cos(\theta)d\theta$ $du=-\sin(\theta)d\theta\;&space;\;&space;\;&space;\;&space;\;&space;v=\sin(\theta)$ $\sin(\theta)\cos(\theta)+\int&space;\sin^2(\theta)d\theta=\sin(\theta)\cos(\theta)+\int\left&space;(&space;1-\cos^2(\theta)&space;\right&space;)d\theta=$ $\sin(\theta)\cos(\theta)+\int&space;d\theta-\int\cos^2(\theta)d\theta=\sin(\theta)\cos(\theta)+\theta-\int\cos^2(\theta)d\theta$

As we saw for the integral of sin(x)cos(x) here https://addjustabitofpi.com/integration-by-parts-solution-with-procedure/ we notice that

$\int&space;\cos^2(\theta)d\theta=\sin(\theta)\cos(\theta)+\theta-\int\cos^2(\theta)d\theta$

This is exactly like an equation, so we can more the integral on the right to the left and we’ll get

$2\int&space;\cos^2(\theta)d\theta=\sin(\theta)\cos(\theta)+\theta$

And so

$\int&space;\cos^2(\theta)d\theta=\frac{\sin(\theta)\cos(\theta)+\theta}{2}$

That’s it! The “+C” is missing because we have yet to undo the substitution. Remember that x=sin(theta), so theta=arcsin(x).

$\frac{\sin(\arcsin(x))\cos(\theta)+\arcsin(x)}{2}=\frac{x\cos(\theta)+\arcsin(x)}{2}$

As you can see, I haven’t replaced the theta inside cos. This is because we don’t want to write cos(arcsin(x)). Although it is mathematically correct, it doesn’t make much sense. Instead, we can think if cos(theta) as sqrt(1-sin²(theta)) and here we can replace the theta with arcsin(x), since sin and arcsin cancel out.

$\cos(\theta)=\sqrt{1-\sin^2(\theta)}=\sqrt{1-\sin^2(\arcsin(x))}=\sqrt{1-x^2}$

The result will therefore be

$\int&space;\sqrt{1-x^2}dx=\frac{x\sqrt{1-x^2}+\arcsin(x)}{2}+C$

What an integral! I will give you more examples in another post, involving tangent, secant and more! Stay tuned! And if you have any questions, leave a comment, I will be happy to help you!

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