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, \[f(x)=\tan \frac{x}{2}\] \[\Rightarrow \] \[f'(x)={{\sec }^{2}}\frac{x}{2}\cdot \frac{1}{2}=\frac{1}{2}{{\sec }^{2}}\frac{x}{2}\] \[\therefore \] \[l={{e}^{x}}f(x)+C={{e}^{x}}\tan \frac{x}{2}+C\] \[[\because \,\,\,\int{{{e}^{x}}[f(x)+f'(x)}]dx={{e}^{x}}f(x)+C\]
You need to login to perform this action.
You will be redirected in
3 sec