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If $I_n=\int_0^{\pi / 4} \tan ^n \theta d \theta$ for $n=1,2,3, \ldots$, then $I_{n-1}+I_{n+1}$ is equal to
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Verified Answer
The correct answer is:
$\frac{1}{n}$
Given, $I_n=\int_0^{\pi / 4} \tan ^n \theta d \theta$
$\begin{aligned}
& \therefore \quad I_{n-1}+I_{n+1}=\int_0^{\pi / 4} \tan ^{n-1} \theta d \theta \\
& \quad+\int_0^{\pi / 4} \tan ^{n+1} \theta d \theta \\
& =\int_0^{\pi / 4} \tan ^{n-1} \theta\left(1+\tan ^2 \theta\right) d \theta \\
& =\int_0^{\pi / 4} \sec ^2 \theta \tan ^{n-1} \theta d \theta
\end{aligned}$
Put $\tan \theta=\phi \Rightarrow \sec ^2 \theta d \theta=d \phi$
$\begin{aligned}
\therefore \quad I_{n-1}+I_{n+1} & =\int_0^1 \phi^{n-1} d \phi \\
& =\left[\frac{\phi^n}{n}\right]_0^1=\frac{1}{n}
\end{aligned}$
$\begin{aligned}
& \therefore \quad I_{n-1}+I_{n+1}=\int_0^{\pi / 4} \tan ^{n-1} \theta d \theta \\
& \quad+\int_0^{\pi / 4} \tan ^{n+1} \theta d \theta \\
& =\int_0^{\pi / 4} \tan ^{n-1} \theta\left(1+\tan ^2 \theta\right) d \theta \\
& =\int_0^{\pi / 4} \sec ^2 \theta \tan ^{n-1} \theta d \theta
\end{aligned}$
Put $\tan \theta=\phi \Rightarrow \sec ^2 \theta d \theta=d \phi$
$\begin{aligned}
\therefore \quad I_{n-1}+I_{n+1} & =\int_0^1 \phi^{n-1} d \phi \\
& =\left[\frac{\phi^n}{n}\right]_0^1=\frac{1}{n}
\end{aligned}$
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