Hans Georg Schaathun
Oktober 2016
$$\int \frac{1}{1+\sin\theta+\cos \theta}d\theta = $$
$$x=\tan\frac\theta2$$
$$\theta=2\tan^{-1}x$$
$$x=\tan\frac\theta2$$
$$\theta=2\tan^{-1}x$$
$$\cos^2\frac{\theta}2=$$
$$\cos^2\frac{\theta}2 = \frac1{\sec^2\frac\theta2}$$
$$= \frac1{1+\tan^2\frac\theta2}$$
$$\cos^2\frac{\theta}2 = \frac1{1+x^2}$$
$$x=\tan\frac\theta2$$
$$\theta=2\tan^{-1}x$$
$$\cos^2\frac{\theta}2 = \frac1{1+x^2}$$
$$\cos\theta=$$
$$\cos\theta= 2\cos^2\frac\theta2 -1$$
$$= \frac2{1+x^2} - 1 $$
$$\cos\theta= \frac{1-x^2}{1+x^2} $$
$$x=\tan\frac\theta2$$
$$\theta=2\tan^{-1}x$$
$$\cos^2\frac{\theta}2 = \frac1{1+x^2}$$
$$\cos\theta= \frac{1-x^2}{1+x^2} $$
$$\sin\theta=$$
$$\sin\theta= 2\sin\frac\theta2\cos\frac\theta2$$
$$=2\tan\frac\theta2\cos^2\frac\theta2 $$
$$\sin\theta= \frac{2x}{1+x^2} $$
$$x=\tan\frac\theta2$$
$$\theta=2\tan^{-1}x$$
$$\cos^2\frac{\theta}2 = \frac1{1+x^2}$$
$$d\theta = $$
$$\frac{dx}{d\theta} = \frac12\sec^2\frac\theta2$$
$$\frac{d\theta}{dx} = 2\cos^2\frac\theta2 = \frac2{1+x^2} $$
$${d\theta}= \frac2{1+x^2} dx$$
$$\cos\theta= \frac{1-x^2}{1+x^2} $$
$$\sin\theta= \frac{2x}{1+x^2} $$
$${d\theta}= \frac2{1+x^2} dx$$
$$\int \frac{1}{1+\sin\theta+\cos \theta}d\theta $$
$$= \int \frac1 {1+\frac{2x}{1+x^2}+\frac{1-x^2}{1+x^2}} \cdot\frac2{1+x^2} dx$$
$$= \int \frac{2dx} {(1+x^2)+2x+(1-x^2)} $$
$$= \int \frac{2dx}{2+2x} = \int \frac{dx}{1+x} $$
$$= \ln|1+x|+C = \ln|1+\tan\frac\theta2|+C $$