A) \[{{\tan }^{n\,}}x+c\]
B) \[\frac{({{\tan }^{n-1}}x)}{n-1}+c\]
C) \[\frac{{{\tan }^{n}}x}{n}+c\]
D) \[n\,\,{{\tan }^{n}}x+c\]
Correct Answer: B
Solution :
We have, \[{{I}_{n}}=\int{ta{{n}^{n}}x\,dx}\] \[=\int{{{\tan }^{n-2}}\,x({{\tan }^{2}}x)\,dx}\] \[\int{{{\tan }^{n-2}}}x({{\sec }^{2}}x-1)dx\] \[\int{{{\tan }^{n-2}}x.\,\,{{\sec }^{2}}x\,dx-{{I}_{n-2}}}\] \[\Rightarrow \] \[{{I}_{n}}+{{I}_{n-2}}=\int{{{\tan }^{n-2}}}x.\,{{\sec }^{2}}x\,dx\] Put \[\tan \,x=t\] \[\Rightarrow \] \[{{\sec }^{2}}x\,\,dx=dt\] \[{{I}_{n}}+{{I}_{n-2}}=\int{{{t}^{n-2}}dt}\] \[=\frac{{{t}^{n-1}}}{n-1}+c\] \[\Rightarrow \] \[{{I}_{n}}+{{I}_{n-2}}=\frac{(tan{{\,}^{n-1}}x)}{n-1}+c\]
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