\(\int \frac{2 \tan (x)}{1+2 \tan ^2(x)} d x=\)
\(\int \frac{2 \tan (x)}{1+2 \tan ^2(x)} d x=\)
- \(\log \left|\cos ^2 x+\sin ^2 x\right|+c\)
- \(\log \left|\frac{\cos ^2 x}{2}+\sin ^2 x\right|+c\)
- \(\log \left|\cos ^2 x+\frac{\sin ^2 x}{2}\right|+c\)
- \(\log \left|\frac{\cos ^2 x}{2}+\frac{\sin ^2 x}{2}\right|+c\)
Solution
\(\begin{aligned}
& \int\left(\frac{2 \tan x}{1+2 \tan ^2 x}\right) d x=\int \frac{2 \sin x \cdot \cos x}{\cos ^2 x+2 \sin ^2 x} d x \\
& \Rightarrow \quad \int \frac{2 \sin x \cdot \cos x}{1+\sin ^2 x} d x
\end{aligned}\)
Let \(1+\sin ^2 x=t \Rightarrow 2 \sin x \cdot \cos x d x=d t\)
\(\begin{aligned}
\Rightarrow \quad \int \frac{d t}{t} & =\log |t|+c_1=\log \left|1+\sin ^2 x\right|+c_1 \\
& =\log \left|2 \sin ^2 x+\cos ^2 x\right|+c_1
\end{aligned}\)
\(\begin{aligned}
& \log 2\left|\sin ^2 x+\frac{\cos ^2 x}{2}\right|+c_1 \\
& \log 2+\left|\sin ^2 x+\frac{\cos ^2 x}{2}\right|+c_1 \\
& \therefore \log \left|\frac{\cos ^2 x}{2}+\sin ^2 x\right|+c \quad\left[\because \log 2+c_1=c\right]
\end{aligned}\)
Asked in: AP EAMCET 2020 (17 Sep Shift 1)
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