Question

The value of $\int \frac{dx}{(3\sin x-4\cos x{)}^2}$ is
(where $C$ is integration constant)

• A. $-\frac{1}{3}\cdot \frac{1}{(3\tan x-4)}+C$
• B. $-\frac{4}{3}\cdot \frac{1}{(3\cot x-4)}+C$
• C. $-\frac{1}{3}\cdot \frac{1}{(3\cot x-4)}+C$
• D. $-\frac{4}{3}\cdot \frac{1}{(3\tan x-4)}+C$

Answer 1

The correct option is A $-\frac{1}{3}\cdot \frac{1}{(3\tan x-4)}+C$

$I=\int \frac{dx}{(3\sin x-4\cos x{)}^2}$
Dividing numerator and denominator by ${\cos }^{2}x$
$=\int \frac{{\sec }^{2}xdx}{(3\tan x-4{)}^2}$

Put $3\tan x-4=t$
$\Rightarrow 3{\sec }^{2}xdx=dt$
$\Rightarrow I=\int \frac{\frac{dt}{3}}{t^2}=-\frac{1}{3}\cdot \frac{1}{(3\tan x-4)}+C$