A) \[\frac{{{\sec }^{5}}x}{5}+\frac{{{\sec }^{3}}x}{3}+c\]
B) \[\frac{{{\tan }^{5}}x}{5}+\frac{{{\tan }^{3}}x}{3}+c\]
C) \[\frac{{{\tan }^{5}}x}{5}+\frac{{{\sec }^{3}}x}{3}+c\]
D) \[\frac{{{\sec }^{5}}x}{5}+\frac{{{\tan }^{3}}x}{3}+c\]
Correct Answer: B
Solution :
[b] Let \[I=\int{{{\tan }^{2}}x{{\sec }^{4}}xdx}\] Let \[\tan x=t\] \[\Rightarrow {{\sec }^{2}}xdx=dt\] \[\therefore I=\int{{{\tan }^{2}}x.{{\sec }^{2}}x.{{\sec }^{2}}x.dx}\] \[=\int{{{\tan }^{2}}x(1+{{\tan }^{2}}x){{\sec }^{2}}x.dx}\] \[\therefore I=\int{{{t}^{2}}(1+{{t}^{2}})dt=\int{({{t}^{2}}+{{t}^{4}})dt}}\] \[=\frac{{{t}^{5}}}{5}+\frac{{{t}^{3}}}{3}+c=\frac{{{\tan }^{5}}x}{5}+\frac{{{\tan }^{3}}x}{3}+c\]
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