Search any question & find its solution
Question:
Answered & Verified by Expert
$\int_{0}^{\pi / 2} \frac{\cos x-\sin x}{1+\cos x \sin x} d x$ is equal to
Options:
Solution:
2660 Upvotes
Verified Answer
The correct answer is:
$0$
Let $I=\int_{0}^{\pi / 2} \frac{\cos x-\sin x}{1+\cos x \sin x} d x$
$\ldots$ (i)
$$
\begin{array}{r}
=\int_{0}^{\pi / 2} \frac{\cos \left(\frac{\pi}{2}-x\right)-\sin \left(\frac{\pi}{2}-x\right)}{1+\cos \left(\frac{\pi}{2}-x\right) \sin \left(\frac{\pi}{2}-x\right)} d x \\
{\left[\because \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right]}
\end{array}
$$
$$
\therefore \quad I=\int_{0}^{\pi / 2} \frac{\sin x-\cos x}{1+\sin x \cos x} d x
$$
On adding Eqs. (i) and (ii), we get
$$
2 I=0 \Rightarrow I=0
$$
$\ldots$ (i)
$$
\begin{array}{r}
=\int_{0}^{\pi / 2} \frac{\cos \left(\frac{\pi}{2}-x\right)-\sin \left(\frac{\pi}{2}-x\right)}{1+\cos \left(\frac{\pi}{2}-x\right) \sin \left(\frac{\pi}{2}-x\right)} d x \\
{\left[\because \int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right]}
\end{array}
$$
$$
\therefore \quad I=\int_{0}^{\pi / 2} \frac{\sin x-\cos x}{1+\sin x \cos x} d x
$$
On adding Eqs. (i) and (ii), we get
$$
2 I=0 \Rightarrow I=0
$$
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.