Q115 / 160 · AP EAMCET 2018 (23 Apr Shift 2) · Mathematics

$$ \int\left(\sec ^4 x+\tan ^4 x\right) d x= $$

  • A
    $\frac{2}{3} \tan ^3 x-\frac{2}{3} \tan x+x+c$
  • B
    $\frac{1}{3} \sec ^2 x \tan x+\frac{5}{3} \tan x+\frac{\tan ^3 x}{3}+x+c$
  • Correct answer
    C
    $\frac{2}{3} \tan ^3 x+x+c$
  • D
    $\frac{1}{3} \sec ^2 x \tan x-\frac{5}{3} \tan x+\frac{\tan ^3 x}{3}+x+c$

Solution

$\begin{aligned} & \text { } I=\int\left(\sec ^4 x+\tan ^4 x\right) d x \\ & =\int\left[\sec ^4 x+\left(\sec ^2 x-1\right)^2\right] d x \\ & =\int\left(2 \sec ^4 x-2 \sec ^2 x+1\right) d x \\ & =\int\left[2\left(1+\tan ^2 x\right) \sec ^2 x-2 \sec ^2 x+1\right] d x \\ & =\int\left(2 \tan ^2 x \sec ^2 x+1\right) d x=\frac{2}{3} \tan ^3 x+x+c .\end{aligned}$

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

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