Integrals — MATHS STD 12 Science — Question

$=\int {\frac{{dx}}{{\sqrt {{{\sin }^4}x.\frac{{\sin (x + \alpha )}}{{\sin x}}} }}} $

$=\int {\frac{{dx}}{{{{\sin }^2}x\sqrt {\frac{{\sin (x + \alpha )}}{{\sin x}}} }} = \int {\frac{{\ cose{c^2}dx}}{{\sqrt {\frac{{\sin (x + \alpha )}}{{\sin x}}} }}} } $

$=\int {\frac{{\ cose{c^2}xdx}}{{\sqrt {\frac{{\sin x.\cos \alpha + \cos x.\sin \alpha }}{{\sin x}}} }}} $
$=\int {\frac{{\cos e{c^2}2dx}}{{\sqrt {\cos \alpha + \cot x.\sin \alpha } }}} $

Put $\cos \alpha + \cot x.\sin \alpha = t$

$0 - \ cose{c^2}x.\sin \alpha dx = dt$

$\ cose{c^2}xdx = \frac{{ - dt}}{{\sin \alpha }}$

$\therefore$$ I= \frac{{ - 1}}{{\sin \alpha }}\int {\frac{{dt}}{{\sqrt t }} = \frac{{ - 1}}{{\sin \alpha }}} \times \frac{{{t^{1/2}}}}{{1/2}} + c$
$ = \frac{{ - 2\sqrt t }}{{\sin \alpha }} + c$

$ = \frac{{ - 2\sqrt {\cos \alpha + \cot x\sin \alpha } }}{{\sin \alpha }} + c$