Evaluate the integral: int frac{dx}{1 - sin x | vedclass

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Question

(A) To solve the integral $I = \int \frac{dx}{1 - \sin x}$, we multiply the numerator and denominator by $(1 + \sin x)$:$I = \int \frac{1 + \sin x}{(1

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$	an x + \sec x + c$

Vedclass · Answer

(A) To solve the integral $I = \int \frac{dx}{1 - \sin x}$, we multiply the numerator and denominator by $(1 + \sin x)$:$I = \int \frac{1 + \sin x}{(1 - \sin x)(1 + \sin x)} dx$$I = \int \frac{1 + \sin x}{1 - \sin^2 x} dx$Using the identity $1 - \sin^2 x = \cos^2 x$, we get:$I = \int \frac{1 + \sin x}{\cos^2 x} dx$$I = \int (\sec^2 x + \sec x 	an x) dx$Integrating term by term:$I = 	an x + \sec x + c$.

Evaluate the integral: $\int \frac{dx}{1 - \sin x}$

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