First slide

Introduction to integration

Question

If $\int \frac{\cos 4 x + 1}{\cot x - \tan x} = K \cos 4 x + C,$ then

Easy

A

$K = - 1 / 2$
B

$K = - 1 / 8$
C

$K = - 1 / 5$
D

none of these

Solution

$\int \frac{\cos 4 x + 1}{\cot x - \tan x} d x$
$= \int \frac{2 \cos^{2} 2 x}{\cos^{2} x - \sin^{2} x} \cdot \sin x \cos x d x$
$= \int \cos 2 x \sin 2 x d x = \frac{1}{2} \int \sin 4 x d x$
$= - \frac{1}{8} \cos 4 x + C$
Hence $K = - 1 / 8$

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