Q115 / 160 · AP EAMCET 2018 (22 Apr Shift 1) · Mathematics

For $n \geq 2$, If $I_n=\int \sec ^n x d x$, then $I_4-\frac{2}{3} I_2=$

  • A
    $\sec ^2 x \tan x+c$
  • Correct answer
    B
    $\frac{1}{3} \sec ^2 x \tan x+c$
  • C
    $\frac{2}{3} \sec ^2 x \tan x+c$
  • D
    $\frac{1}{2} \log |\sec x+\tan x|+c$

Solution

We have, $$ \begin{aligned} & \qquad I_n=\int \sec ^n x d x \\ & \therefore \quad I_2=\int \sec ^2 x d x=\tan x+c_1 \\ & \text { and } \quad I_4=\int \sec ^4 x d x \\ & =\int \sec ^2 x \cdot \sec ^2 x d x \\ & =\int\left(\tan ^2 x+1\right) \sec ^2 x d x \\ & =\frac{\tan ^3 x}{3}+\tan x+c_2 \\ & \therefore \quad=\frac{1}{3} \tan ^3 x+\frac{1}{3} \tan x+c \\ & \quad \frac{I_4}{3} I_2=\frac{\tan ^3 x}{3}+\tan x+c_2-\frac{2}{3} \tan x-\frac{2 c_1}{3} \\ & \quad=\frac{1}{3} \tan x\left(\tan ^2 x+1\right)+c \\ & =\frac{1}{3} \tan x \sec ^2 x+c \end{aligned} $$

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

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