Integral of $\frac{\sin x}{1+\sin^2x}$ from 0 to $\pi/2$

Question

I am trying yo find $\int_0^{\pi/2}\frac{\sin x}{1+\sin^2x}dx$.

So far I have tried using the substitution $\tan u=\sin x$ which led me to $$\int_{u=0}^{u=\pi/4}\frac{\sin x}{\cos x}du$$

$$\int_{0}^{\pi/4}\frac{\tan u}{\sqrt{1+\tan^2u}}du$$ which I initially thought would allow me to rearrange to

$$\int_{0}^{\pi/4}\frac{1}{\sqrt{\cot^2u+1}}du$$ so that I could use the result

$$\int\frac{1}{\sqrt{x^2+a^2}}du=ln(x+\sqrt{x^2+a^2})$$ However I realised that this would not work because when I divided by tanu II divided by zero as one of my limits is u=0...

I would be very grateful for any hints and also if anyone has any tips for integrating tricky trig functions like this one. I really would like to develop y integrating toolbox so that I can tackle integrals like this in the future...

Thank you in advance :)

Answer 1

$$ \int_{0}^{\pi/2}\frac{\sin x}{1+\sin^2x}\,dx= \int_{0}^{\pi/2}\frac{\sin x}{2-\cos^2x}\,dx $$ Do $t=\cos x$, so $dt=-\sin x\,dx$ and the integral becomes $$ \int_{1}^{0}-\frac{1}{2-t^2}\,dt $$

Answer 2

HINT:

Use $\sin^2x=1-\cos^2x$

Set $\cos x=-u$

See this