Indefinte Integration

∫dx/tanx+cotx+secx+cosecx

1 Answers

1708
man111 singh ·

\hspace{-16}\bf{=\int\frac{1}{(\tan x+\cot x)+(\sec x+\csc x)}dx}$\\\\\\ $\bf{=\int\frac{1}{\frac{\sin^2 x+\cos ^2 x}{\sin x.\cos x}+\frac{\sin x+\cos x}{\sin x.\cos x}}dx}$\\\\\\ $\bf{=\int\frac{\sin x.\cos x}{1+(\sin x+\cos x)}dx}$\\\\\\ $\bf{=\frac{1}{2}\int \frac{\sin 2x}{(\sin x+\cos x)+1}dx}$\\\\\\ $\bf{=\frac{1}{2}\int \frac{\sin 2x.\left\{(\sin x+\cos x)-1\right\}}{\left\{(\sin x+\cos x)+1\right\}\cdot\left\{(\sin x+\cos x)-1\right\}}dx}$\\\\\\ $\bf{=\frac{1}{2}\int\frac{\sin 2x.\left\{(\sin x+\cos x)-1\right\}}{(\sin x+\cos x)^2-1}dx}$\\\\\\ $\bf{=\frac{1}{2}\int \left\{\sin x+\cos x-1\right\}dx}$\\\\\\ $\bf{=\frac{1}{2}\left\{-\cos x+\sin x-x\right\}+\mathbb{C}}$

Your Answer

Close [X]