Question 115 of 246MediumMCQ
\[{I_n} = \int\limits_0^{\pi /4} {{{\tan }^n}x\,dx} \] then \[\,\mathop {\lim }\limits_{n \to \infty } \,n\left[ {{I_n} + {I_{n + 2}}} \right]\] equals
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Official solution
In+In+2
\[ = \int\limits_0^{\pi /4} {{{\tan }^n}} \,x\left( {1 + {{\tan }^2}x} \right)dx\]
\[ = \int\limits_0^{\pi /4} {{{\tan }^n}} x\,{\sec ^2}x\,dx\]
\[ = \left[ {{{{{\tan }^{n + 1}}x} \over {n + 1}}} \right]_0^{\pi /4}\]
\[ = {{1 - 0} \over {n + 1}} = {1 \over {n + 1}}\]
\[\therefore\] \[{I_n} + {I_{n + 2}} = {1 \over {n + 1}}\]
\[ \Rightarrow \mathop {\lim }\limits_{n \to \infty } n\left[ {{I_n} + {I_{n + 2}}} \right]\]
\[ = \mathop {\lim }\limits_{n \to \infty } \,n.{1 \over {n + 1}} = \mathop {\lim }\limits_{n \to \infty } {n \over {n + 1}}\]
\[ = \mathop {\lim }\limits_{n \to \infty } {n \over {n\left( {1 + {1 \over n}} \right)}} = 1\] Answer the question to unlock