Question

$\int_{}^{}\frac{\cos 4x - 1}{\cot x - \tan x}$dx is equal to –

• A. –1/2 cos 4x + c
• B. –1/4 cos 4x + c
• C. –1/2 sin 2x + c
• D. None of these

Answer 1

Answer: (D)

None of these

Answer 2

$\int \frac { \cos 4 x - 1 } { \cot x - \tan x }$dx =$\int_{}^{}\frac{- 2\sin^{2}2x}{\frac{\cos 2x}{\sin x\cos x}}$dx

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

Put cos 2x = t Ž – sin 2x dx = $\frac{dt}{2}$

= $\frac { 1 } { 2 }$ $\int_{}^{}\frac{1 - t^{2}}{t}$dt = $\frac{1}{2}$log t – $\frac{t^{2}}{4}$ + c

= $\frac{1}{2}$log cos 2x – $\frac{\cos^{2}2x}{4}$ + c.