A) \[\frac{{{\sec }^{p+1}}x}{p+1}+c\]
B) \[\frac{{{\sec }^{p}}x}{p}+c\]
C) \[\frac{{{\tan }^{p+1}}x}{p+1}+c\]
D) \[\frac{{{\tan }^{p}}x}{p}+c\]
Correct Answer: B
Solution :
Put \[\sec x=t\Rightarrow \sec x\tan x\,dx=dt,\] therefore \[\int_{{}}^{{}}{{{\sec }^{p}}x\tan x\,dx}=\int_{{}}^{{}}{{{t}^{p-1}}dt=\frac{{{t}^{p}}}{p}+c}=\frac{{{\sec }^{p}}x}{p}+c.\]
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