Integral Calculus Question 379
Question: The value of $ \int_{{}}^{{}}{\frac{\sin x}{{{\cos }^{2}}x}\ dx} $ is
Options:
A) $ \sin x+k $
B) $ \tan x+k $
C) $ \sec x+k $
D) $ \tan x+\sec x+k $
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Answer:
Correct Answer: C
Solution:
Given $ I=\int_{{}}^{{}}{\frac{\sin x}{{{\cos }^{2}}x}},dx $ . Put $ \cos x=t\Rightarrow \sin x,dx=-dt $
$ \therefore ,I=\int_{{}}^{{}}{\frac{-dt}{t^{2}}}=\frac{1}{t}+k=\sec x+k $ .
BETA
