A) \[x+\cos x+c\]
B) \[1+\sin x+c\]
C) \[\sec x-\tan x+c\]
D) \[\sec x+\tan x+c\]
Correct Answer: D
Solution :
\[\int_{{}}^{{}}{\frac{dx}{1-\sin x}=\int_{{}}^{{}}{\frac{(1+\sin x)}{1-{{\sin }^{2}}x}\,dx}}\] \[=\int_{{}}^{{}}{{{\sec }^{2}}x\,dx+\int_{{}}^{{}}{\tan x\,.\,\sec x\,dx}}\]\[=\tan x+\sec x+c\].
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