91Ó°ÊÓ

In order to simplify the integrand, we can multiply both the numerator and the denominator by the conjugate of the denominator $$(1-\sin\theta)$$. This results in: $$\int \frac{1}{1+\sin \theta} d\theta = \int \frac{1-\sin \theta}{1-\sin^2 \theta} d\theta$$ Recall the trigonometric identity: $$1 - \sin^2\theta= \cos^2\theta$$ Now, the integral becomes: $$\int \frac{1-\sin \theta}{\cos^2 \theta} d\theta$$

Let $$u = \cos{\theta}$$ So, $$\frac{du}{d\theta} = -\sin{\theta}$$ Further, $$d\theta = \frac{du}{-\sin \theta}$$ Now, substitute these into the integral: $$\int \frac{1-\sin\theta}{\cos^2\theta} d\theta = \int \frac{1-\sin\theta}{u^2} \frac{du}{-\sin\theta}$$ Simplify the above equation: $$-\int \frac{1-\sin\theta}{u^2} du$$

Now, we need to change the remaining $$\sin\theta$$ to a function of $$u$$. We know that: $$u = \cos{\theta}$$ By using the Pythagorean identity, we have: $$1 - \sin^2{\theta} = \cos^2{\theta} = u^2$$ So, $$\sin{\theta} = \sqrt{1-u^2}$$ Now, substitute this back into the integral: $$-\int \frac{1-\sqrt{1-u^2}}{u^2} du$$

Split the integral into two parts: $$-\int \frac{1}{u^2} du + \int \frac{\sqrt{1-u^2}}{u^2} du$$ The first integral can be solved directly: $$-\int \frac{1}{u^2} du = \frac{1}{u} + C_1$$ For the second integral, we can make use of a known integral for the arcsin function. Recall the following integral: $$\int \frac{1}{\sqrt{1-u^2}} du = \arcsin u + C_2$$ Now, to get the integral of $$\frac{\sqrt{1-u^2}}{u^2}$$. Observe that: $$\frac{d}{du} (\arcsin u) = \frac{1}{\sqrt{1-u^2}}$$ Thus, by integration by parts, let: $$dv = \frac{\sqrt{1-u^2}}{u^2} du$$ And, $$u = \arcsin u$$ So, $$du = \frac{1}{\sqrt{1-u^2}} du$$ $$v = -\frac{1}{u}$$ Now, apply integration by parts: $$\int \frac{\sqrt{1-u^2}}{u^2} du = -\frac{\arcsin u}{u} - \int -\frac{1}{u} \frac{1}{\sqrt{1-u^2}} du = -\frac{\arcsin u}{u} + C_3$$

Now, combine the results: $$-\int \frac{1-\sin\theta}{\cos^2\theta} d\theta = \frac{1}{u} - \frac{\arcsin u}{u} + C$$ Finally, substitute back the original variable using $$u = \cos\theta$$: $$\int \frac{d \theta}{1+\sin \theta} = \frac{1}{\cos\theta} - \frac{\arcsin (\cos\theta)}{\cos\theta} + C$$