Q114 / 160 · TS EAMCET 2022 (20 Jul Shift 2) · Mathematics

If $x \neq(2 n+1) \frac{\pi}{2}, n \in \mathbb{Z}$ and $\cos x \neq \frac{-1}{2}$ then $$ \int\left(\frac{\sin x+\sin 2 x}{1+\cos x+\cos 2 x}\right)^2 d x= $$

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

Solution

We have $\int\left(\frac{\sin (x)+\sin (2 x)}{1+\cos (x)+\cos (2 x)}\right)^2 d x$ $$ \begin{aligned} & \Rightarrow \int\left(\frac{\sin (x)(1+2 \cos (x)}{\cos (x)(1+2 \cos (x)}\right)^2 \\ & \Rightarrow \int\left(\sec ^2 x-1\right) d x \\ & \Rightarrow \tan (x)-x+c \end{aligned} $$

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

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