Question

The value of the definite integral $\underset{0}{\overset{\frac{3\pi }{4}}{\int }}\left(\right(1+x)\sin x+(1-x\left)\cos x\right)dx,$ is

• A. $2\tan \left(\frac{3\pi }{8}\right)$
• B. $2\tan \left(\frac{\pi }{8}\right)$
• C. $2\cot \left(\frac{3\pi }{8}\right)$
• D. $2\cot \left(\frac{\pi }{8}\right)$

Answer 1

Answer: (A), (D)

The correct options are
A $2\tan \left(\frac{3\pi }{8}\right)$
D $2\cot \left(\frac{\pi }{8}\right)$

$I=\underset{0}{\overset{\frac{3\pi }{4}}{\int }}(\sin x+\cos x)dx+\underset{0}{\overset{\frac{3\pi }{4}}{\int }}x\cdot (\sin x-\cos x)dx$
$=\underset{0}{\overset{\frac{3\pi }{4}}{\int }}(\sin x+\cos x)dx+\left[x\right(-\cos x-\sin x){]}_0^{\frac{3\pi }{4}}+\underset{0}{\overset{\frac{3\pi }{4}}{\int }}(\sin x+\cos x)dx$
$=2\underset{0}{\overset{\frac{3\pi }{4}}{\int }}(\sin x+\cos x)dx$

$=2(\sqrt{2}+1)$