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If $I_{n}=\int_{0}^{\frac{\pi}{4}} \tan ^{n} x d x$, where $n$ is positive integer, then $I_{10}+I_{8}$ is equal to
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Verified Answer
The correct answer is:
$\frac{1}{9}$
$I_{n}=\int_{0}^{\frac{\pi}{4}} \tan ^{n} x d x...(i)$
$I_{n+2}=\int_{0}^{\frac{\pi}{4}} \tan ^{n+2} x d x...(ii)$
Adding Eqs. (i) and (ii),
$\begin{aligned}
I_{n}+I_{n+2} &=\int_{0}^{\frac{\pi}{4}} \tan ^{n} x d x+\int_{0}^{\frac{\pi}{4}} \tan ^{n+2} x d x \\
&=\int_{0}^{\frac{\pi}{4}} \tan ^{n} x \cdot \sec ^{2} x d x \\
&=\left[\frac{\tan ^{n+1} x}{n+1}\right]_{0}^{\frac{\pi}{4}}=\frac{1}{n+1}
\end{aligned}$
Thus, $I_{n}+I_{n+2}=\frac{1}{(n+1)}$
Substitute 8 for $x$,
$I_{8}+I_{10}=\frac{1}{9}$
$I_{n+2}=\int_{0}^{\frac{\pi}{4}} \tan ^{n+2} x d x...(ii)$
Adding Eqs. (i) and (ii),
$\begin{aligned}
I_{n}+I_{n+2} &=\int_{0}^{\frac{\pi}{4}} \tan ^{n} x d x+\int_{0}^{\frac{\pi}{4}} \tan ^{n+2} x d x \\
&=\int_{0}^{\frac{\pi}{4}} \tan ^{n} x \cdot \sec ^{2} x d x \\
&=\left[\frac{\tan ^{n+1} x}{n+1}\right]_{0}^{\frac{\pi}{4}}=\frac{1}{n+1}
\end{aligned}$
Thus, $I_{n}+I_{n+2}=\frac{1}{(n+1)}$
Substitute 8 for $x$,
$I_{8}+I_{10}=\frac{1}{9}$
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