Definite Integration Question 402
Question: $ \int\limits _{\pi /4}^{3\pi /4}{\frac{dx}{1+\cos x}} $ is equal to
[IIT 1999]
Options:
A) 2
B) $ -2 $
C) $ \frac{1}{2} $
D) $ -\frac{1}{2} $
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Answer:
Correct Answer: A
Solution:
$ \int _{\pi /4}^{3\pi /4}{\frac{dx}{1+\cos x}} $
$ \int _{\pi /4}^{3\pi /4}{\frac{1-\cos x}{1-{{\cos }^{2}}x}}dx=\int _{\pi /4}^{3\pi /4}{\frac{1-\cos x}{{{\sin }^{2}}x}}dx $
$ =\int _{\pi /4}^{3\pi /4}{(cose{c^{2}}x}-\cot xcosecx)dx $
$ =(-\cot x+cosecx\text{)} _{\pi \text{/4}}^{3\pi \text{/4}}=2 $ .
BETA
