Question

The function $f(\theta )=\frac{d}{d\theta }{\int }_0^{\theta }\frac{dx}{1-\cos \theta \cos x}$ satisfies the differential equation

• A. $\frac{df}{d\theta }+2f\left(\theta \right)\cot \theta =0$
• B. $\frac{df}{d\theta }-2f\left(\theta \right)\cot \theta =0$
• C. $\frac{df}{d\theta }+2f\left(\theta \right)=0$
• D. $\frac{df}{d\theta }-2f\left(\theta \right)=0$

Answer 1

Answer: (A)

$\frac{df}{d\theta }+2f\left(\theta \right)\cot \theta =0$

$f\left(\theta \right)=\frac{d}{d\theta }{\int }_0^{\theta }\frac{dx}{1-cosxcos\theta }$
$\Rightarrow f\left(\theta \right)=\frac{d\theta }{d\theta }\left[\frac{1}{1-co{s}^2\theta }\right]=cose{c}^2\theta$
$\Rightarrow \frac{df}{d\theta }=2cosec\theta (-cosec\theta cot\theta )$
$\Rightarrow \frac{df}{d\theta }=-2cose{c}^2\theta cot\theta$
$\Rightarrow \frac{df}{d\theta }=-2f\left(\theta \right)cot\theta$
$\Rightarrow \frac{df\left(\theta \right)}{d\theta }+2f\left(\theta \right)cot\theta =0$