Integral Calculus Question 230
Question: $ \int_{{}}^{{}}{\frac{dx}{1-\sin x}}= $
[MP PET 1991]
Options:
A) $ x+\cos x+c $
B) $ 1+\sin x+c $
C) $ \sec x-\tan x+c $
D) $ \sec x+\tan x+c $
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Answer:
Correct Answer: D
Solution:
$ \int_{{}}^{{}}{\frac{dx}{1-\sin x}=\int_{{}}^{{}}{\frac{(1+\sin x)}{1-{{\sin }^{2}}x},dx}} $ $ =\int_{{}}^{{}}{{{\sec }^{2}}x,dx+\int_{{}}^{{}}{\tan x,.,\sec x,dx}} $ $ =\tan x+\sec x+c $ .
BETA
