Question

$\int (\tan x+2\tan 2x+4\tan 4x+8\cot 8x)dx$ has the value

• A. $\ln \left(\sin x\right)+c$
• B. $-\ln \left(\text{cosec}x\right)+c$
• C. $\ln \left(\cos x\right)+c$
• D. $\ln \left(\sin 2x\right)-\ln \left(2\cos x\right)+c$

Answer 1

Answer: (A), (B)

The correct options are
A $\ln \left(\sin x\right)+c$
B $-\ln \left(\text{cosec}x\right)+c$

Let $I=\int (\tan x+2\tan 2x+4\tan 4x+8\cot 8x)dx$

$\tan x-\cot x=\tan x-\frac{1}{\tan x}$

$=\left(\frac{{\tan }^{2}x-1}{\tan x}\right)\frac{\times 2}{\times 2}$

$=\frac{2}{-\tan 2x}=-2\cot 2x$

Back to integral, adding and subtracting $\cot x$, we get

$I=\int (\tan x-\cot x+2\tan 2x+4\tan 4x+8\cot 8x+\cot x)dx$

$=\int (-2\cot 2x+2\tan 2x+4\tan 4x+8\cot 8x+\cot x)dx$

$=\int \cot x$

Now, we have to integrate $\cot x$

$I=\int \cot x$

$=\ln \sin x+c=-\ln \left(\text{cosec}x\right)+c$