#### Provide Solution For R.D. Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise  Revision Exercise Question 78 Maths Textbook Solution.

$\ln |\sec x+\tan x|-\sin x-\frac{\sin 3 x}{3}-\frac{\sin ^{5} x}{5}+c$

Hint:

To solve the given equation we have to split the sin? x term in sin? x.sin² x.

Given:

$\int \frac{\sin ^{6} x}{\cos x} d x$

Solution:

$I=\int \frac{\sin ^{6} x}{\cos x} d x$

$=\int \frac{\sin ^{4} x \sin ^{2} x}{\cos x} d x=\int \frac{\sin ^{4} x}{\cos x}\left(1-\cos ^{2} x\right) d x$

$=\int \frac{\sin ^{4} x}{\cos x} d x-\int \frac{\sin ^{4} x \cos ^{2} x}{\cos x} d x$

$=\int \frac{\sin ^{4} x}{\cos x} d x-\int \sin ^{4} x \cos x d x$

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

$=\int \frac{\sin ^{2} x}{\cos x} d x-\int \sin ^{2} x \cos x d x-\frac{t^{5}}{5}$

$=\int \frac{\left(1-\cos ^{2} x\right)}{\cos x} d x-\int t^{2} d t-\frac{\sin ^{5} x}{5}$

$=\int \frac{1}{\cos x} d x-\int \cos x d x-\frac{t^{8}}{3}-\frac{\sin ^{5} x}{5}$

$=\int \sec x d x-\sin x-\frac{\sin 3 x}{3}-\frac{\sin ^{5} x}{5}$

$=\ln |\sec x+\tan x|+\left(-\sin x-\frac{\sin 3 x}{3}-\frac{\sin ^{5} x}{5}\right)+c$

$=\ln |\sec x+\tan x|-\sin x-\frac{\sin 3 x}{3}-\frac{\sin ^{5} x}{5}+c$