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If $I_{n}=\int_{0}^{\frac{\pi}{4}} \tan ^{n} x d x$ then what is $I_{n}+I_{n-2}$ equal to?
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Verified Answer
The correct answer is:
$\frac{1}{(n-1)}$
Let $I_{n}=\int_{0}^{\pi / 4} \tan ^{n} x d x$
Consider,
$I_{n}+I_{n-2}=\int_{0}^{\pi / 4} \tan ^{n} x d x+\int_{0}^{\pi / 4} \tan ^{n-2} x d x$
$=\int_{0}^{\pi / 4} \tan ^{n-2} x\left(\tan ^{2} x+1\right) d x$
$=\int_{0}^{\pi / 4} \sec ^{2} x \tan ^{n-2} x d x$
Put $\tan x=t$
$\sec ^{2} x d x=d t$
when $x=0$ then $t=0$ and when $x=\frac{\pi}{4}, t=1$
$\therefore I_{n}+I_{n-2}=\int_{0}^{1} t^{n-2} d t$
$=\left.\frac{t^{n-2+1}}{n-2+1}\right|_{0} ^{1}=\left.\frac{t^{n-1}}{n-1}\right|_{0} ^{1}=\frac{1}{n-1}[1-0]=\frac{1}{n-1}$
Consider,
$I_{n}+I_{n-2}=\int_{0}^{\pi / 4} \tan ^{n} x d x+\int_{0}^{\pi / 4} \tan ^{n-2} x d x$
$=\int_{0}^{\pi / 4} \tan ^{n-2} x\left(\tan ^{2} x+1\right) d x$
$=\int_{0}^{\pi / 4} \sec ^{2} x \tan ^{n-2} x d x$
Put $\tan x=t$
$\sec ^{2} x d x=d t$
when $x=0$ then $t=0$ and when $x=\frac{\pi}{4}, t=1$
$\therefore I_{n}+I_{n-2}=\int_{0}^{1} t^{n-2} d t$
$=\left.\frac{t^{n-2+1}}{n-2+1}\right|_{0} ^{1}=\left.\frac{t^{n-1}}{n-1}\right|_{0} ^{1}=\frac{1}{n-1}[1-0]=\frac{1}{n-1}$
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