Q86 / 160 · AP EAMCET 2010 · Mathematics

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

  • A
    arithmetic progression
  • B
    geometric progression
  • Correct answer
    C
    harmonic progression
  • D
    arithmetico-geometric progression

Solution

$I_n=\int_0^{\pi / 4} \tan ^n x d x$ 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.

Hence, the correct answer is option: 'C'.

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