First slide

Introduction to integration

Question

If $\int \frac{\cos 4 x + 1}{\cot x - \tan x} dx = Acos 4 x + B$ then

Moderate

A

$A = - \frac{1}{2}$
B

$A = - \frac{1}{8}$
C

$A = - \frac{1}{4}$
D

None of these

Solution

$\int \frac{\cos 4 x + 1}{\cot x - \tan x} dx = Acos 4 x + B$
Let,
$\begin{array}{l} I = \int \frac{\cos 4 x + 1}{\cot x - \tan x} dx \\ = \int \frac{2 \cos^{2} 2 x}{\frac{\cos x}{\sin x} - \frac{\sin x}{\cos x}} dx = \int \frac{2 \cos^{2} 2 x}{\frac{\cos 2 x}{\sin xcos x}} dx \\ = \int \sin 2 xcos 2 xdx \\ = \frac{1}{2} \int 2 \sin 2 xcos 2 xdx \\ = \frac{1}{2} \int \sin 4 xdx = \frac{1}{2} [- \frac{\cos 4 x}{4}] + B \\ = - \frac{1}{8} \cos 4 x + B \\ but I = Acos 4 x + B (given) \\ ∴ A = - \frac{1}{8} \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App