Q111 / 160 · TS EAMCET 2020 (10 Sep Shift 2) · Mathematics
For $x \in\left(\frac{3 \pi}{4}, \pi\right), \int(\sqrt{1+\sin 2 x}+\sqrt{1-\sin 2 x})$ $d x=$
- Correct answerA$-2 \cos x+C$
- B$2 \sin x+C$
- C$-2 \sin x+c$
- D$2 \cos x+C$
Solution
$\left.\int(\sqrt{1+\sin 2 x})+\sqrt{1-\sin 2 x}\right) d x$
$=\int\left(\sqrt{\sin ^2 x+\cos ^2 x+2 \sin x \cos x}\right.$ $+\sqrt{\left.\sin ^2 x+\cos ^2 x-2 \sin x \cos x\right)} d x$
$=\int((\sin x+\cos x)+(\sin x-\cos x)) d x$ $\left(\because x \in\left(\frac{3 \pi}{4}, \pi\right)\right)$
$=\int 2 \sin x d x=2 \int \sin x d x$
$=2(-\cos x)+C=-2 \cos x+C$
Hence, the correct answer is option: 'A'.