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AP EAPCET 25-Aug-2021 Shift 2 Paper

Section: Mathematics

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Question : 76 of 160

Marks: +1, -0

If

\int\frac{1+\cos(4x)}{\cot(x)-\tan(x)}dx=A\cos(4x)+B

A=

______

$-\frac{1}{2}$
$-\frac{1}{4}$
$-\frac{1}{3}$
$-\frac{1}{8}$

Solution: 👈: Video Solution

\int\frac{1+\cos(4x)}{\cos x-\tan x}dx=A\cos(4x)+B

, then

A=?

Let

I=\int\frac{1+\cos(2\times 2x)}{\cot x-\tan x}dx

=\int\frac{1+2\cos^{2}(2x)-1}{\frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}}dx

=\int\frac{2\cos^{2}(2x)}{\cos^{2}x-\sin^{2}x}\cdot(\sin x\cdot\cos x)dx

I=2\int\frac{\cos^{2}(2x)}{\cos(2x)}(\sin x\cdot\cos x)dx

I=\int\cos(2x)\sin(2x)dx

=\frac{1}{2}\int2\sin(2x)\cos(2x)dx=\frac{1}{2}\int\sin(4x)dx

I=\frac{-\cos(4x)}{8}+C=k\cos(4x)+C

∴

k=-\frac{1}{8}

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