Question

If $\int \frac{dx}{\sin x{\cos }^{3}x}=\ln |f\left(x\right)|+g\left(x\right)+C$, then the number of solution(s) of the equation $g\left(x\right)-f\left(x\right)=0$ in $[0,(2n+1\left)\frac{\pi }{2}\right],(n\in W)$ is

• A. $(n+1)$
• B. $(2n+1)$
• C. $2(n+1)$
• D. $n$

Answer 1

The correct option is C $2(n+1)$

$I=\int \frac{dx}{\sin x{\cos }^{3}x}=\int \frac{dx}{\frac{\sin x}{\cos x}\times {\cos }^{4}x}=\int \frac{{\sec }^{4}xdx}{\tan x}$
Substituting $\tan x=z$
$\Rightarrow {\sec }^{2}xdx=dz$
$\therefore I=\int \left(\frac{1+z^2}{z}\right)dz=\int (\frac{1}{z}+z)dz$
$=\ln |z|+\frac{z^2}{2}+C=\ln |\tan x|+\frac{{\tan }^{2}x}{2}+C$
$\Rightarrow f\left(x\right)=\tan x,g\left(x\right)=\frac{{\tan }^{2}x}{2}$
$\therefore$ the equation $g\left(x\right)-f\left(x\right)=0$
$\Rightarrow \frac{{\tan }^{2}x}{2}-\tan x=0$
$\Rightarrow \tan x(\tan x-2)=0\Rightarrow \tan x=0,\tan x=2$
number of solutions for $\tan x=0$ in $[0,(2n+1\left)\frac{\pi }{2}\right]$
$=(n+1)$
number of solutions for $\tan x=2$ in $[0,(2n+1\left)\frac{\pi }{2}\right]=(n+1)$
Total number of solutions $=2(n+1)$