Question

$\text{If}\int \frac{(3\sin x+4)}{(3+4\sin x{)}^2}=f\left(x\right)+C,\text{where}f\left(0\right)=-\frac{1}{3}\text{and}f\left(\frac{\pi }{2}\right)=0\text{(}C\text{is constant of integration), then}$

• A. The minimum value of |$f\left(x\right)$| is $\frac{1}{3}$
• B. The minimum value of |$f\left(x\right)$| is 0
• C. The value of $f\left(\pi \right)$ is $\frac{1}{3}$
• D. The value of $f\left(\pi \right)$ is 0

Answer 1

Answer: (B), (C)

The correct options are
B The minimum value of |$f\left(x\right)$| is 0
C The value of $f\left(\pi \right)$ is $\frac{1}{3}$

Multiply and divide by ${\sec }^{2}x$
$\int \frac{3\tan x\sec x+4{\sec }^{2}x}{(3\sec x+4\tan x{)}^2}dx$
$3\sec x+4\tan x=t$
$(3\sec x\tan x+4{\sec }^{2}x)dx=dt$
$\int \frac{dt}{t^2}=-\frac{1}{(3\sec x+4\tan x)}+c$
$f\left(0\right)=-\frac{1}{3}\text{and}f\left(\frac{\pi }{2}\right)=0$
$f\left(x\right)=-\frac{\cos x}{3+4\sin x}|f\left(x\right){|}_{min}=0\text{and}f\left(\pi \right)=\frac{1}{3}$