Evaluate the following integrals:


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,

$=\log |\sin x|-\frac{\sin ^{6} x}{6}-\frac{3 \sin ^{2} x}{2}+\frac{3 \sin ^{4} x}{4}+C$

$\therefore \int \frac{\cos ^{7} x}{\sin x} d x=\log |\sin x|-\frac{\sin ^{6} x}{6}-\frac{3 \sin ^{2} x}{2}+\frac{3 \sin ^{4} x}{4}+C$

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