Integral Calculus Question 61
Question: $ \int_{{}}^{{}}{\frac{dx}{\cos x-\sin x}} $ is equal to
[AIEEE 2004]
Options:
A) $ \frac{1}{\sqrt{2}}\log | \tan ( \frac{x}{2}+\frac{3\pi }{8} ), |+c $
B) $ \frac{1}{\sqrt{2}}\log | \cot ( \frac{x}{2} ), |+c $
C) $ \frac{1}{\sqrt{2}}\log | \tan ( \frac{x}{2}-\frac{3\pi }{8} ), |+c $
D) $ \frac{1}{\sqrt{2}}\log | \tan ( \frac{x}{2}-\frac{\pi }{8} ), |+c $
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Answer:
Correct Answer: A
Solution:
We have, $ I=\int_{{}}^{{}}{\frac{dx}{\cos x-\sin x}=\frac{1}{\sqrt{2}}\int_{{}}^{{}}{\frac{d^{2}}{\cos ( \frac{\pi }{4}+x )}}} $ $ I=\frac{1}{\sqrt{2}}\int_{{}}^{{}}{\sec ( x+\frac{\pi }{4} )},dx=\frac{1}{\sqrt{2}}\log | \tan ( \frac{\pi }{4}+\frac{x}{2}+\frac{\pi }{8} ), |+c $ $ I=\frac{1}{\sqrt{2}}\log | \tan ( \frac{x}{2}+\frac{3\pi }{8} ), |+c $ .
BETA
