Question

The value of $\int \frac{\sin 7x}{\cos x}dx$ is
(where $C$ is constant of integration)

• A. $-\frac{\cos 6x}{3}+\frac{2\cos 4x}{3}-\cos 2x+\cos x+C$
• B. $-\frac{\cos 6x}{3}+\frac{\cos 4x}{2}-\cos 2x+\ln |\cos x|+C$
• C. $-\frac{\sin 6x}{3}+\frac{2\sin 3x}{3}-\sin 2x+\ln |\sin x|+C$
• D. $-\frac{\sin 6x}{3}+2\sin 4x-\sin 2x+\sin x+C$

Answer 1

The correct option is B $-\frac{\cos 6x}{3}+\frac{\cos 4x}{2}-\cos 2x+\ln |\cos x|+C$

Let
$I=\int \frac{\sin 7x}{\cos x}dx=\int \frac{(\sin 7x+\sin 5x)-(\sin 5x+\sin 3x)+(\sin 3x+\sin x)-\sin x}{\cos x}dx=\int \frac{2(\sin 6x-\sin 4x+\sin 2x)\cos x-\sin x}{\cos x}dx=\int \left[2\right(\sin 6x-\sin 4x+\sin 2x)-\tan x]dx=2[-\frac{\cos 6x}{6}+\frac{\cos 4x}{4}-\frac{\cos 2x}{2}]-\ln |\sec x|+C=-\frac{\cos 6x}{3}+\frac{\cos 4x}{2}-\cos 2x+\ln |\cos x|+C$