## Trigonometric Indefinite Integral

jacks on Wednesday, August 12 2015, 03:55 AM
0

Evaluation of $$\displaystyle \int\frac{1}{\csc x+\cos x}dx$$

•
Replied by jacks on Thursday, November 05 2015, 04:45 AM · Hide · #1
Let $$\displaystyle I = \int\frac{1}{\sin x+\sec x} dx = \int\frac{\cos x}{\sin x\cdot \cos x+1} =\int \frac{2\cos x}{\sin 2x+2}dx$$

So Integral $$\displaystyle I = \int \frac{(\cos x+\sin x)+(\cos x-\sin x)}{\sin 2x+2}dx$$

$$\displaystyle = \int\frac{\sin x+\cos x}{3-(\sin x-\cos x)^2}dx+\int \frac{\cos x-\sin x}{(\sin x+\cos x)^2+1}dx$$

In first Integral Put $$(\sin x-\cos x) = t\;,$$ Then $$(\cos x+\sin x)dx = dt$$

and in second Integral Put $$(\sin x+\cos x) = u\;,$$ Then $$(\cos x-\sin x)dx = du$$

So $$\displaystyle I = \int\frac{1}{\left(\sqrt{3}\right)^2-t^2}dt+\int \frac{1}{1+u^2}du = \frac{1}{2\sqrt{3}}\cdot \ln\left|\frac{\sqrt{3}+t}{\sqrt{3}-t}\right|+\tan^{-1}(u)+\mathcal{C}$$

So $$\displaystyle I =\int\frac{1}{\sin x+\sec x} dx = \frac{1}{2\sqrt{3}}\cdot \ln\left|\frac{\sqrt{3}+(\sin x-\cos x)}{\sqrt{3}-(\sin x-\cos x)}\right|+\tan^{-1}(\sin x+\cos x)+\mathcal{C}$$