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Can someone please help me to integrate this function $$\int_0^1 \frac{1}{\sqrt{1-x^2}} \frac{s}{s^2+\alpha^2 \cdot x^2} dx = \frac{s\pi \sqrt{\frac{\alpha^2}{s^2}}}{2\alpha^2\sqrt{1+\frac{s^2}{\alpha^2}}}$$ My professor to me to substitute $x = \sin u$ to get $$s \int_0^{\pi/2} \frac{1}{s^2+\alpha^2 \cdot \sin^2 u} du$$ But I can't seem to get where the $\frac{1}{\sqrt{1-x^2}}$ go

Isn't it supposed to be $\frac{1}{\sqrt{1-\sin^2 u}}$. Or $\sec u$

Please help

2 Answers 2

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Notice that we have the pythagorean identity $\sin^2u+\cos^2u=1$, or $\cos^2u=1-\sin^2u$. That is how the first fraction was handled. The remaining can be dealt with $\sin^2u=\frac{1-\cos(2u)}2$ and a tangent half-angle substitution, as described in this post.

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Remember also the substitution law. You need to change $x$ to $\sin u$, but it is the only thing to change?





SPOILER:

Remember also to change $dx$ to $du$, that is: $$ dx=\sin'udu. $$

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    I don't think you hit any part of the question. Please reread, as the OP has already tried the substitution... – 2017-02-09
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    @user414565 I completely forgot about it. Now I get it. Thank you – 2017-02-09