Evaluate
$$
\int \frac{\sin^{3/2}(a)+\cos^{3/2}(a)}{\sqrt{\sin^3(a)\cos^3(a)\sin(a+b)}}\,da
$$
My try: rearranging original integral
$$
\int \frac{{1+\tan^{3/2}}(a)}{\sqrt{\sin^3(a)\sin(a+b)}}\,da
= \int \frac{{(1+\tan^{3/2}}(a))\sec^2(a)}{\sqrt{\tan^4(a)\cos(b)+\tan^3(a)\sin(b)}}\,da
$$
let $t=\tan a$ then it becomes
$$
\int \frac{{1+t^{3/2}}}{\sqrt{t^4\cos b+t^3\sin b}}\,dt
$$
now I am stuck.could anyone give me any hints to proceed further or any simpler method
Best Answer
Hint:
$\int \frac{\sin^{3/2}(a)+\cos^{3/2}(a)}{\sqrt{\sin^3(a)\cos^3(a)\sin(a+b)}}\,da$ = $\int \frac{1}{\sqrt{\sin^3(a)\sin(a+b)}}\,da + \int \frac{1}{\sqrt{\cos^3(a)\sin(a+b)}}\,da$
Take first part now.
$\sqrt{\sin^3(a)\sin(a+b)} = \sqrt{\sin^4(a)(\cos b + \cot a \sin b)}$
say, $t = \cos b + \cot a \sin b$
$dt = - \dfrac {1}{\sin^2(a)} \sin b \,da$
So, first integral $\int \frac{1}{\sqrt{\sin^3(a)\sin(a+b)}}\,da = -\dfrac{1}{\sin b} \int \frac{1}{\sqrt t} \,dt$
You can do the second part similarly. Can you take it from here?