First slide

Methods of integration

Question

$If \int \frac{\cos 4 x + 1}{\cot x - \tan x} d x = A \cos 4 x + B, then$

Easy

A

$A = - 1 / 2$
B

$A = - 1 / 8$
C

$A = - 1 / 4$
D

$None of these$

Solution

$\begin{array}{l} \int \frac{\cos 4 x + 1}{\cot x - \tan x} d x = \int \frac{2 \cos^{2} 2 x}{\cos^{2} x - \sin^{2} x} \sin x \cos x d x \\ = \int \cos 2 x \sin 2 x d x \\ = \frac{1}{4} \int \sin 4 x d x = - \frac{1}{8} \cos 4 x + C \end{array}$
$Hence, A = - 1 / 8 and B \in R$

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