Evaluate $\int \frac{\cos ^{7} x}{\sin x} d x$
$\int \frac{\cos ^{7} x}{\sin x} d x$
We can write above integral as:
$\int \frac{\left(\cos ^{2} x\right)^{3} \cdot \cos x}{\sin x} d x^{-(1)}$
Put $\operatorname{Sin} x=t$
Differentiting w.r.t $x$ we get,
$\cos x \cdot d x=d t$
$\therefore$ integral $(1)$ becomes,
$=\int \frac{\left(\cos ^{2} x\right)^{3}}{t} d t$
$=\int \frac{\left(1-\sin ^{2} x\right)^{2}}{t} d t \cdots\left(\because \sin ^{2}(x)+\cos ^{2}(x)=1\right)$
$=\int \frac{\left(1-t^{2}\right)^{3}}{t} d t$
$=\int \frac{(1)^{3}-\left(t^{2}\right)^{3}-3(1)\left(t^{2}\right)\left(1-t^{2}\right)}{t} d t=\int \frac{1-t^{6}-3 t^{2}+3 t^{4}}{t} d t$
$=\int \frac{1}{t} d t-\int \frac{t^{6}}{t} d t-\int \frac{3 t^{2}}{t} d t+\int \frac{3 t^{4}}{t} d t$
$=\log |t|-\frac{t^{6}}{6}-\frac{3 t^{2}}{2}+\frac{3 t^{4}}{4}+C$
Putting value of $t=\sin (x)$ we get,
$=\log |\sin x|-\frac{\sin ^{6} x}{6}-\frac{3 \sin ^{2} x}{2}+\frac{3 \sin ^{4} x}{4}+C$
$\therefore \int \frac{\cos ^{7} x}{\sin x} d x=\log |\sin x|-\frac{\sin ^{6} x}{6}-\frac{3 \sin ^{2} x}{2}+\frac{3 \sin ^{4} x}{4}+C$