Integrate $\int^{\frac{\pi}{2}}_0 \frac{a+b\sin x}{(b+a\sin x)^2}dx$

Question

Integrate

$$\int^{\frac{\pi}{2}}_ 0 \frac{a+b\sin x}{(b+a\sin x)^2}dx$$

(note: answer is $1/b$)

I have no idea where to start with this, is there meant to be a simple way to integrate this?

Thanks so much everyone in advance .

Answer 1

Hint

Divide both the numerator and denominator by $\cos ^2 x$ to change integral to $$\int_0^{\frac {\pi}{2}} \frac {a\sec^2 x+b\sec x\tan x}{(b\sec x+a\tan x)^2}dx$$

Make the substitution $$b\sec x+a\tan x=u$$ and hence $$du=( a\sec^2 x+b\sec x\tan x)dx$$ to get the answer as $\frac 1b$

Answer 2

Try the so-called Weierstrass substitution: $$\sin(x)=\frac{2t}{1+t^2}$$ and $$dx=\frac{2t}{1+t^2}dt$$ and your integrand will get the form $${\frac { \left( {t}^{2}+1 \right) \left( a{t}^{2}+2\,bt+a \right) }{ \left( b{t}^{2}+2\,at+b \right) ^{2}}} $$ and your indefinite integral will be $$\int\frac{(at^2+2bt+a)2t}{(bt^2+2at+b)^2}dt$$