Let \(f(x)=\tan ^{-1}\left(\frac{1+\cos x}{\sin x}\right)\); \(g(x)=\tan ^{-1}\left(\frac{\sin x}{1-\cos…
Let \(f(x)=\tan ^{-1}\left(\frac{1+\cos x}{\sin x}\right)\); \(g(x)=\tan ^{-1}\left(\frac{\sin x}{1-\cos x}\right)\), then \(\int(f(x)+g(x)) d x=\)
- \(\frac{\pi x}{2}-\frac{x^2}{4}\)
- \(\pi x-\frac{x^2}{2}\)
- \(\pi x+\frac{x^2}{4}\)
- \(\pi x+\frac{x^2}{2}\)
Solution
\(\begin{aligned}
f(x) & =\tan ^{-1}\left(\frac{1+\cos x}{\sin x}\right) \\
& =\tan ^{-1}\left(\frac{2 \cos ^2 \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}}\right)=\tan ^{-1}\left(\cot \frac{x}{2}\right) \\
& =\tan ^{-1}\left[\tan \left(\frac{\pi}{2}-\frac{x}{2}\right)\right]=\frac{\pi}{2}-\frac{x}{2} \\
\text { and } g(x) & =\tan ^{-1}\left(\frac{\sin x}{1-\cos x}\right) \\
& =\tan ^{-1}\left(\frac{2 \sin x / 2 \cos x / 2}{2 \sin x / 2}\right)=\tan ^{-1}\left(\cot \frac{x}{2}\right) \\
& =\tan ^{-1}\left[\tan \left(\frac{\pi}{2}-\frac{x}{2}\right)\right]=\frac{\pi}{2}-\frac{x}{2} \\
\therefore \int(f(x) & +g(x)) d x=\int(\pi-x) d x=\pi x-\frac{x^2}{2}+c
\end{aligned}\)
Hence, option (b) is correct.
Asked in: AP EAMCET 2020 (18 Sep Shift 2)
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