A) \[\sin x+k\]
B) \[\tan x+k\]
C) \[\sec x+k\]
D) \[\tan x+\sec x+k\]
Correct Answer: C
Solution :
Given \[I=\int_{{}}^{{}}{\frac{\sin x}{{{\cos }^{2}}x}}\,dx\]. Put \[\cos x=t\Rightarrow \sin x\,dx=-dt\] \[\therefore \,\,\,I=\int_{{}}^{{}}{\frac{-dt}{{{t}^{2}}}}=\frac{1}{t}+k=\sec x+k\].You need to login to perform this action.
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