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Answered & Verified by Expert
If $x \neq(2 n+1) \frac{\pi}{2}, n \in \mathbb{Z}$ and $\cos x \neq \frac{-1}{2}$ then
$$
\int\left(\frac{\sin x+\sin 2 x}{1+\cos x+\cos 2 x}\right)^2 d x=
$$
Options:
$$
\int\left(\frac{\sin x+\sin 2 x}{1+\cos x+\cos 2 x}\right)^2 d x=
$$
Solution:
2182 Upvotes
Verified Answer
The correct answer is:
$\tan x-x+c$
We have $\int\left(\frac{\sin (x)+\sin (2 x)}{1+\cos (x)+\cos (2 x)}\right)^2 d x$
$$
\begin{aligned}
& \Rightarrow \int\left(\frac{\sin (x)(1+2 \cos (x)}{\cos (x)(1+2 \cos (x)}\right)^2 \\
& \Rightarrow \int\left(\sec ^2 x-1\right) d x \\
& \Rightarrow \tan (x)-x+c
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow \int\left(\frac{\sin (x)(1+2 \cos (x)}{\cos (x)(1+2 \cos (x)}\right)^2 \\
& \Rightarrow \int\left(\sec ^2 x-1\right) d x \\
& \Rightarrow \tan (x)-x+c
\end{aligned}
$$
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