For -frac{pi}{2}

Question

(D) Given the integral $I = \int 	an^{-1} \left( \sqrt{\frac{1 - \sin x}{1 + \sin x}} ight) dx$. Using the identities $1 - \sin x = (\cos \frac{x}{

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$\frac{\pi}{4} x - \frac{x^2}{4} + C$

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(D) Given the integral $I = \int 	an^{-1} \left( \sqrt{\frac{1 - \sin x}{1 + \sin x}} ight) dx$. Using the identities $1 - \sin x = (\cos \frac{x}{2} - \sin \frac{x}{2})^2$ and $1 + \sin x = (\cos \frac{x}{2} + \sin \frac{x}{2})^2$,we have: $I = \int 	an^{-1} \sqrt{\frac{(\cos \frac{x}{2} - \sin \frac{x}{2})^2}{(\cos \frac{x}{2} + \sin \frac{x}{2})^2}} dx$ $I = \int 	an^{-1} \left( \frac{\cos \frac{x}{2} - \sin \frac{x}{2}}{\cos \frac{x}{2} + \sin \frac{x}{2}} ight) dx$ Dividing numerator and denominator by $\cos \frac{x}{2}$,we get: $I = \int 	an^{-1} \left( \frac{1 - 	an \frac{x}{2}}{1 + 	an \frac{x}{2}} ight) dx$ Using the formula $	an(\frac{\pi}{4} - 	heta) = \frac{1 - 	an 	heta}{1 + 	an 	heta}$: $I = \int 	an^{-1} \left( 	an \left( \frac{\pi}{4} - \frac{x}{2} ight) ight) dx$ $I = \int \left( \frac{\pi}{4} - \frac{x}{2} ight) dx$ Integrating term by term: $I = \frac{\pi}{4} x - \frac{x^2}{4} + C$.

For $-\frac{\pi}{2} < x < \frac{\pi}{2}$,evaluate the integral $\int \tan^{-1} \left( \sqrt{\frac{1 - \sin x}{1 + \sin x}} \right) dx$ (where $C$ is a constant of integration).

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