Question

Evaluate the following integrals:

$\int \frac{1}{\cos x+\mathrm{cosec}x}\mathrm{d}x$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{1}{\cos x+\mathrm{cosec}x}\mathrm{d}x \\ =\int \frac{\sin x}{1+\cos x\sin x}\mathrm{d}x \\ =\int \frac{2\sin x}{2+2\cos x\sin x}\mathrm{d}x \\ =\int \frac{\sin x+\cos x+\sin x-\cos x}{2+2\cos x\sin x}\mathrm{d}x \\ =\int \frac{\sin x+\cos x}{2+2\cos x\sin x}\mathrm{d}x+\int \frac{\sin x-\cos x}{2+2\cos x\sin x}\mathrm{d}x \\ =\int \frac{\sin x+\cos x}{3-{\sin }^{2}x-{\cos }^{2}x+2\cos x\sin x}\mathrm{d}x+\int \frac{\sin x-\cos x}{1+{\sin }^{2}x+{\cos }^{2}x+2\cos x\sin x}\mathrm{d}x \\ =\int \frac{\sin x+\cos x}{3-{\left(\sin x-\cos x\right)}^2}\mathrm{d}x+\int \frac{\sin x-\cos x}{1+{\left(\sin x+\cos x\right)}^2}\mathrm{d}x \\ =I_1+I_2...\left(1\right) \\ \mathrm{where},I_1=\int \frac{\sin x+\cos x}{3-{\left(\sin x-\cos x\right)}^2}\mathrm{d}x\mathrm{and}{I}_2=\int \frac{\sin x-\cos x}{1+{\left(\sin x+\cos x\right)}^2}\mathrm{d}x \\ \mathrm{Now}, \\ {I}_1=\int \frac{\sin x+\cos x}{3-{\left(\sin x-\cos x\right)}^2}\mathrm{d}x \\ \mathrm{Let}\left(\sin x-\cos x\right)=t \\ \mathrm{On}\mathrm{differentiating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \left(\cos x+\sin x\right)\mathrm{d}x=\mathrm{d}t \\ \therefore I_1=\int \frac{1}{3-{\left(t\right)}^2}\mathrm{d}t \\ =\frac{1}{2\sqrt{3}}\log \left|\frac{\sqrt{3}+t}{\sqrt{3}-t}\right|+c_1 \\ =\frac{1}{2\sqrt{3}}\log \left|\frac{\sqrt{3}+\sin x-\cos x}{\sqrt{3}-\left(\sin x-\cos x\right)}\right|+c_1...\left(2\right) \\ \mathrm{Now}, \\ {I}_2=\int \frac{\sin x-\cos x}{1+{\left(\sin x+\cos x\right)}^2}\mathrm{d}x \\ \mathrm{Let}\left(\sin x+\cos x\right)=t \\ \mathrm{On}\mathrm{differentiating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \left(\cos x-\sin x\right)\mathrm{d}x=\mathrm{d}t \\ \therefore I_2=-\int \frac{1}{1+{\left(t\right)}^2}\mathrm{d}t \\ =-{\tan }^{-1}t+c_2 \\ =-{\tan }^{-1}\left(\sin x+\cos x\right)+c_2...\left(3\right) \\ \mathrm{On}\mathrm{substituting}\left(2\right)\mathrm{and}\left(3\right)\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ I=\frac{1}{2\sqrt{3}}\log \left|\frac{\sqrt{3}+\sin x-\cos x}{\sqrt{3}-\sin x+\cos x}\right|-{\tan }^{-1}\left(\sin x+\cos x\right)+c \\ \mathrm{Hence},\int \frac{1}{\cos x+\mathrm{cosec}x}\mathrm{d}x=\frac{1}{2\sqrt{3}}\log \left|\frac{\sqrt{3}+\sin x-\cos x}{\sqrt{3}-\sin x+\cos x}\right|-{\tan }^{-1}\left(\sin x+\cos x\right)+c \end{gathered}$$