Provide Solution For  R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise  Revision Exercise Question 19  Maths Textbook Solution.

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

Given:

$\int \frac{\cos ^{7} x}{\sin x} d x$

Hint:

Use trigonometric identities and substitution method of integration.

Solution:

$\int \frac{\cos ^{7} x}{\sin x} d x$

Let $t=\sin x$

$d t=\cos x d x$                            $(on differentiating)$

$now,$

$I=\int \frac{\cos ^{7} x}{\sin x} d x$

$=\int \frac{\left(1-\sin ^{2} t\right)\left(1-\sin ^{2} t\right)\left(1-\sin ^{2} t\right)}{t} d t$

$=\int \frac{\left(1-t^{2}\right)^{2}\left(1-t^{2}\right)}{t} d t$

$=\int \frac{\left(1+t^{4}-2 t^{2}\right)\left(1-t^{2}\right)}{t} d t$

$=\int \frac{1-3 t^{2}+3 t^{4}-t^{6}}{t} d t$

$=\int \frac{1}{t}-3 t+3 t^{3}-t^{5} d t$

$=\operatorname{logt}-\frac{3 t^{2}}{2}+\frac{3 t^{4}}{4}-\frac{t^{6}}{6}+c$

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