# Thread: Integration with a squareroot

x=sin(t)

3. ## Re: Integration with a squareroot

Start by substituting x = sin(w), so that dx = cos(w) dw. You'll end up having to integrate $\cos^2(x)dx$, and to do that you can use the half angle formula: $\cos^2w = \frac 1 2 (1+\cos(2w))$. Try it, and post back with what you get.

4. ## Re: Integration with a squareroot

You may want to check this page which has detailed instructions(and demonstration) on how to integrate $\int \sqrt{1 - x^2}\,\,dx$ as a definite integral. For indefinite integral the procedure is same.

definite integrals

5. ## Re: Integration with a squareroot

To make it easier I re-edited the post. We have to evaluate:
$\int \sqrt{1-x^2}\,\,dx$

Here it is again:

Let's think of $x$ and $1$ in the above question as two sides of a right triangle.

$\cos{(\theta)} = \frac{x}{1} = x$

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

$\text{And } dx = -\sin{(\theta)}\,\,d\theta$

\begin{align*}\int \sqrt{1-x^2}\,\,dx =& -\int \sin^2{(\theta)}\,\,d\theta \\=& \int \frac{\cos{(2\theta)}-1}{2}\,\,d\theta....\text{[Using trigonometry]} \\=& \frac{\sin{(2\theta)}}{4} -\frac{\theta}{2} + C \\=& \frac{\sin{(\theta)}\cos{(\theta)}}{2} - \frac{\theta}{2} + C......\text{[using trigonometry]}\\=& \frac{x\sqrt{1-x^2}}{2} - \frac{\cos^{-1}{(x)}}{2}+ C\\& \text{.....[by replacing } \sin{(\theta)}, \cos{(\theta)} \text{ and } \theta \text{]}\end{align*}

Hope it helps.

6. ## Re: Integration with a squareroot

Note that either x= sin(w), as suggested by ebaines, or $x= sin(\theta)$, as suggested by X3bnm, will work in exactly the same way.