Question

$\int_{}^{}\frac{\sin x}{\sin 4x}dx$ = A log $\left| \frac{1 + \sin x}{1 - \sin x} \right|$ + B log $\left| \frac{1 + \sqrt{2}\sin x}{1 - \sqrt{2}\sin x} \right| + C$

• A. A = $\frac{1}{8}$, B = $\frac{1}{4\sqrt{2}}$
• B. A = – $\frac{1}{8}$, B = – $\frac{1}{4\sqrt{2}}$
• C. A = –$\frac{1}{8}$, B = $\frac{1}{4\sqrt{2}}$
• D. A = $\frac{1}{8}$, B = – $\frac{1}{4\sqrt{2}}$

Answer 1

Answer: (C)

A = –$\frac{1}{8}$, B = $\frac{1}{4\sqrt{2}}$

Answer 2

=

$\int \frac { \sin x d x } { 4 \cos x \sin x \cos 2 x }$= $\frac { 1 } { 4 } \int \frac { \cos x d x } { \cos ^ { 2 } x \cos 2 x }$