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If $I_n=\int_0^{\pi / 4} \tan ^n x d x$, then $I_2+I_4, I_3+I_5, I_4+I_6, \ldots$, are in
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The correct answer is:
harmonic progression
$I_n=\int_0^{\pi / 4} \tan ^n x d x$
We have, $I_{r+2}=\int_0^{\pi / 4} \tan ^{r+2} x d x$
$=\int_0^{\pi / 4} \tan ^r x \cdot \tan ^2 x d x$
and $\quad I_r=\int_0^{\pi / 4} \tan ^r x d x$
Then, $\quad I_r+I_{r+2}=\int_0^{\pi / 4} \tan ^r x d x$ $+\int_0^{\pi / 4} \tan ^r x \cdot \tan ^2 x d x$
$=\int_0^{\pi / 4} \tan ^r x\left(1+\tan ^2 x\right) d x$
$=\int_0^{\pi / 4} \tan ^r x \cdot \sec ^2 x d x$
[Put $t=\tan x \Rightarrow d t=\sec ^2 x d x$ ]
$=\int_0^1 t^r d t \Rightarrow\left[\frac{t^{r+1}}{r+1}\right]_0^1$
$=\left(\frac{1}{r+1}\right)$
So, $\quad I_r+I_{r+2}=\frac{1}{r+1}$
$I_2+I_4=\frac{1}{3}$
$I_3+I_5=\frac{1}{4}$
$I_4+I_6=\frac{1}{5}$
which are clearly in HP.
We have, $I_{r+2}=\int_0^{\pi / 4} \tan ^{r+2} x d x$
$=\int_0^{\pi / 4} \tan ^r x \cdot \tan ^2 x d x$
and $\quad I_r=\int_0^{\pi / 4} \tan ^r x d x$
Then, $\quad I_r+I_{r+2}=\int_0^{\pi / 4} \tan ^r x d x$ $+\int_0^{\pi / 4} \tan ^r x \cdot \tan ^2 x d x$
$=\int_0^{\pi / 4} \tan ^r x\left(1+\tan ^2 x\right) d x$
$=\int_0^{\pi / 4} \tan ^r x \cdot \sec ^2 x d x$
[Put $t=\tan x \Rightarrow d t=\sec ^2 x d x$ ]
$=\int_0^1 t^r d t \Rightarrow\left[\frac{t^{r+1}}{r+1}\right]_0^1$
$=\left(\frac{1}{r+1}\right)$
So, $\quad I_r+I_{r+2}=\frac{1}{r+1}$
$I_2+I_4=\frac{1}{3}$
$I_3+I_5=\frac{1}{4}$
$I_4+I_6=\frac{1}{5}$
which are clearly in HP.
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