Question

The value of $\int \frac{dx}{{\cos }^{2}x(4+\tan x{)}^4}$ is

• A. $\frac{1}{3(4+\tan x{)}^3}+C$
• B. $-\frac{1}{3(4+\tan x{)}^3}+C$
• C. $\frac{1}{5(4+\tan x{)}^5}+C$
• D. $-\frac{1}{5(4+\tan x{)}^5}+C$

Answer 1

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

$I=\int \frac{dx}{{\cos }^{2}x(4+\tan x{)}^4}$
Putting $4+\tan x=t,$ we get
${\sec }^{2}xdx=dt$
$\Rightarrow I=\int \frac{dt}{t^4}=\int t^{-4}dt=\frac{t^{-3}}{-3}+C=-\frac{1}{3t^3}+C$
$\Rightarrow I=-\frac{1}{3(4+\tan x{)}^3}+C$