Q117 / 180 · COMEDK 2016 · Mathematics
If $I_{1}=\int_{0}^{\pi / 2} x \sin x d x$ and $I_{2}=\int_{0}^{\pi / 2} x \cos x d x$, then which one of the following is true?
- A$I_{1}=I_{2}$
- B$I_{1}+I_{2}=0$
- C$I_{1}=\frac{\pi}{2} I_{2}$
- Correct answerD$I_{1}+I_{2}=\frac{\pi}{2}$
Solution
$I_{1}-\int_{0}^{\pi / 2} x \sin x d x$
$=\left[-x \cos x+\int \cos x\right]_{0}^{\pi / 2}$
$=[-x \cos x+\sin x]_{0}^{\pi / 2}$
$=0+\sin \frac{\pi}{2}-0=1$
Similarly, $I_{2}=\int_{0}^{\pi / 2} x \cos x d x$
$\quad=\left[x \sin x-\int_{\sin x]_{0}^{\pi / 2}}\right.$
$\quad=[x \sin x+\cos x]_{0}^{\pi / 2}$
$\quad=\frac{\pi}{2} \sin \frac{\pi}{2}-1=\frac{\pi}{2}-1$
Hence, $I_{1}+I_{2}=1+\frac{\pi}{2}-1=\frac{\pi}{2}$
Hence, the correct answer is option: 'D'.