question_answer

Evaluate \[\int{(\sqrt{\tan \,\,x}\,+\,\sqrt{\cot \,x})}\,dx.\]

OR

Evaluate \[\int{{{e}^{x}}\left( \frac{1+\sin x}{1+\cos x} \right)}\,dx.\]

Answer:

Let \[l=\int{(\sqrt{\tan \,x}\,+\sqrt{\cot \,x})\,dx=\int{\left( \frac{\sqrt{\sin x}}{\sqrt{\cos x}}+\frac{\sqrt{\cos x}}{\sqrt{\sin x}} \right)dx}}\]

\[\left[ \because \,\,\,\tan \theta =\frac{\sin \theta }{\cos \theta }\,\,\text{and}\,\,\cot \theta =\frac{\cos \theta }{\sin \theta } \right]\]

\[=\int{\frac{\sin x+\cos x}{\sqrt{\sin x}\sqrt{\cos x}}dx}\]

\[=\int{\frac{\sin x+\cos x}{\sqrt{\sin x\,\cos x}}dx}=\sqrt{2}\int{\frac{\sin x+\cos x}{\sqrt{2\sin x\,\cos x}}dx}\]\[=\sqrt{2}\int{\frac{\sin x+\cos x}{\sqrt{1-{{(\sin x-\cos x)}^{2}}}}dx}\]

Put \[\sin x-\cos x=t\]

\[\Rightarrow \]   \[(cosx+\sin x)dx=dt\]

\[\therefore \]      \[l=\sqrt{2}\int{\frac{1}{\sqrt{1-{{t}^{2}}}}dt}\]

\[l=\sqrt{2}{{\sin }^{-1}}(t)+C\]  \[\left[ \because \int{\frac{1}{\sqrt{1-{{x}^{2}}}}dx}={{\sin }^{-1}}x+C \right]\]

\[=\sqrt{2}{{\sin }^{-1}}(\sin x-\cos x)+C\] \[\left[ \because t=\sin x-\cos x \right]\]

OR

Let \[l=\int{{{e}^{x}}\left( \frac{1+\sin x}{1+\cos x} \right)}\,dx\]

\[=\int{{{e}^{x}}\left[ \frac{1+2\sin (x/2)\cos (x/2)}{2{{\cos }^{2}}(x/2)} \right]}\,dx\]

­\[\left[ \because \,\,\,\sin \theta =2\sin \frac{\theta }{2}\cos \frac{\theta }{2}\,\,\text{and}\,\,1+\cos \theta =2{{\cos }^{2}}\frac{\theta }{2} \right]\]\[=\int{{{e}^{x}}}\left[ \frac{1}{2{{\cos }^{2}}(x\,/2)}+\frac{2\sin (x/2)\cos (x/2)}{2{{\cos }^{2}}(x\,/2)} \right]\,dx\]\[=\int{{{e}^{x}}}\left[ \frac{1}{2}{{\sec }^{2}}\left( \frac{x}{2} \right)+\tan \frac{x}{2} \right]\,dx\]

Thus, we have integration of the form

\[\int{{{e}^{x}}}\left[ f(x)+f'(x) \right]\,dx\]

Here,     \[f(x)=\tan \frac{x}{2}\]

\[\Rightarrow \]   \[f'(x)={{\sec }^{2}}\frac{x}{2}\cdot \frac{1}{2}=\frac{1}{2}{{\sec }^{2}}\frac{x}{2}\]

\[\therefore \]      \[l={{e}^{x}}f(x)+C={{e}^{x}}\tan \frac{x}{2}+C\]

\[[\because \,\,\,\int{{{e}^{x}}[f(x)+f'(x)}]dx={{e}^{x}}f(x)+C\]