Search any question & find its solution
Question:
Answered & Verified by Expert
The value of the integration $\int_{-\pi / 4}^{\pi / 4}\left(\lambda|\sin x|+\frac{\mu \sin x}{1+\cos x}+\gamma\right) d x$
Options:
Solution:
2962 Upvotes
Verified Answer
The correct answer is:
is independent of $\mu$ only
Let $I=\int_{-\frac{\pi}{4}}^{\pi / 4}\left(\lambda|\sin x|+\frac{\mu \sin x}{1+\cos x}+\gamma\right)$
$\begin{aligned} &=\int_{-\pi / 4}^{\pi / 4}(\lambda|\sin x|+\gamma) d x+\mu \int_{-\frac{\pi}{4}}^{\pi / 4} \frac{\sin x}{1+\cos x} d x \\ \text { Let } f(x)=\frac{\sin x}{1+\cos x} & \\ \Rightarrow f(-x)=\frac{\sin (-x)}{1+\cos (-x)} &=\frac{-\sin x}{1+\cos x}=-f(x) \end{aligned}$
$f(x)$ is an odd function.
$\int_{-\pi}^{\pi / 4}\left(\frac{\sin x}{1+\cos x}\right) d x=0$
$\left.I=\int_{-\frac{\pi}{6}}^{\pi / 4} \lambda|\sin x|+\gamma\right) d x$
$\therefore I$ is independent of $\mu$
$\begin{aligned} &=\int_{-\pi / 4}^{\pi / 4}(\lambda|\sin x|+\gamma) d x+\mu \int_{-\frac{\pi}{4}}^{\pi / 4} \frac{\sin x}{1+\cos x} d x \\ \text { Let } f(x)=\frac{\sin x}{1+\cos x} & \\ \Rightarrow f(-x)=\frac{\sin (-x)}{1+\cos (-x)} &=\frac{-\sin x}{1+\cos x}=-f(x) \end{aligned}$
$f(x)$ is an odd function.
$\int_{-\pi}^{\pi / 4}\left(\frac{\sin x}{1+\cos x}\right) d x=0$
$\left.I=\int_{-\frac{\pi}{6}}^{\pi / 4} \lambda|\sin x|+\gamma\right) d x$
$\therefore I$ is independent of $\mu$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.