If $\int \frac{d x}{1+\sin x}=\tan \left(\frac{x}{2}-\theta\right)+C$, then $\theta=$
If $\int \frac{d x}{1+\sin x}=\tan \left(\frac{x}{2}-\theta\right)+C$, then $\theta=$
- $\frac{\pi}{2}$
- $\frac{\pi}{4}$
- $\pi$
- $\frac{\pi}{6}$
Solution
$\int \frac{d x}{1+\sin x}=\tan \left(\frac{x}{2}-\theta\right)+c$
$
\begin{aligned}
& \text { Now, } \int \frac{d x}{1+\sin x}=\int \frac{1-\sin x}{1-\sin ^2 x} d x+c \\
& =\int \frac{1-\sin x}{\cos ^2 x} d x+c \\
& =\int\left(\sec ^2 x-\sec x \tan x\right) d x+c=\tan x-\sec x+c \\
& =\frac{\sin x}{\cos x}-\frac{1}{\cos x}+c=\frac{\sin x-1}{\cos x}+c \\
& =\frac{-\left(\sin \frac{x}{2}-\cos \frac{x}{2}\right)^2}{\left(\cos ^2 \frac{x}{2}-\sin ^2 \frac{x}{2}\right)}+c=\frac{-\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)}{\cos \frac{x}{2}+\sin \frac{x}{2}} \\
& =-\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)+c=\tan \left(\frac{x}{2}-\frac{\pi}{4}\right)+c
\end{aligned}
$
From $\mathrm{eq}^{\mathrm{n}}$ (i) \& (ii) $\theta=\frac{\pi}{4}$
Asked in: AP EAMCET 2023 (18 May Shift 2)
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