A) \[2\tan \frac{3\pi }{8}\]
B) \[2\tan \frac{\pi }{4}\]
C) \[2\tan \frac{\pi }{8}\]
D) \[0\]
Correct Answer: A
Solution :
\[I=\int\limits_{0}^{3\pi /4}{(\sin x+\cos x)dx}+\int\limits_{0}^{3\pi /4}{\underbrace{x}_{I}}\underbrace{(\sin x-\cos x)}_{II}dx\] \[=\int\limits_{0}^{3\pi /4}{(\sin x+\cos x)dx}+\underbrace{x(-\cos x-\sin x)|_{0}^{3\pi /4}}_{zero}\] \[+\int\limits_{0}^{3\pi /4}{(\sin x+\cos x)}dx\]
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