MCQ
If $I_n=\int_0^{\pi / 4} \tan ^n \theta d \theta$, then $I_8+I_6$ equals
  • A
    $\frac{1}{4}$
  • B
    $\frac{1}{5}$
  • C
    $\frac{1}{6}$
  • D
    $\frac{1}{7}$

Answer

(D)
$I_8+I_6=\int_0^{\frac{\pi}{4}}\left(\tan ^8 \theta+\tan ^6 \theta\right) d \theta$
$=\frac{1}{8-1}=\frac{1}{7}$
$\int_0^{\pi / 4}\left(\tan ^n x+\tan ^{n-2} x\right) d x=\frac{1}{n-1}$

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