question_answer

Evaluate \[\int{\frac{dx}{1-3\sin x}}\]

OR

Evaluate \[\int{\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}}dx.\]

Answer:

\[l=\int{\frac{dx}{1-3\sin x}}\]

\[=\int{\frac{dx}{1-3\left( \frac{2\tan \frac{x}{2}}{1+{{\tan }^{2}}\frac{x}{2}} \right)}}\]  \[\left[ \because \sin x=\frac{2\tan \frac{x}{2}}{1+{{\tan }^{2}}\frac{x}{2}} \right]\]

\[=\int{\frac{dx}{\frac{1-{{\tan }^{2}}\frac{x}{2}-6\tan \frac{x}{2}}{1+{{\tan }^{2}}\frac{x}{2}}}}\]

\[=\int{\frac{{{\sec }^{2}}\frac{x}{2}}{{{\tan }^{2}}\frac{x}{2}-6\tan \frac{x}{2}+1}}\]    \[[\because \,1+ta{{n}^{2}}\theta =se{{c}^{2}}\theta ]\]

Put \[\tan \frac{x}{2}=t\]

On differentiating both sides w.r.t. x, we get

\[{{\sec }^{2}}\frac{x}{2}\cdot \frac{1}{2}=\frac{dt}{dx}\] \[\Rightarrow \] \[{{\sec }^{2}}\frac{x}{2}dx=2\,dt\]

\[\therefore \] \[l=2\int{\frac{dt}{{{t}^{2}}-6t+1}=2\int{\frac{dt}{{{t}^{2}}-6t+1+{{(3)}^{2}}+{{(3)}^{2}}}}}\]

\[=2\int{\frac{dt}{{{(t-3)}^{2}}-8}=2\int{\frac{dt}{{{(t-3)}^{2}}-{{(2\sqrt{2})}^{2}}}}}\]

\[=2\times \frac{1}{2\times 2\sqrt{2}}\,\log \left| \frac{t-3-2\sqrt{2}}{t-3+2\sqrt{2}} \right|+C\]

\[\left[ \therefore \int{\frac{1}{{{x}^{2}}-{{a}^{2}}}dx=\frac{1}{2a}\log \left| \frac{x-a}{x+a} \right|} \right]\]

\[=\frac{1}{2\sqrt{2}}\log \left| \frac{\tan \frac{x}{2}-3-2\sqrt{2}}{\tan \frac{x}{2}-3+2\sqrt{2}} \right|+C\]          \[\left[ \because t=\tan \frac{x}{2} \right]\]

OR

Let        \[l=\int{\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}}dx.\]

Put \[\sqrt{x}=\cos t\] or \[x={{\cos }^{2}}t\]

\[\Rightarrow \] \[dx=-\,2\sin t\,.\,\cos t.dt\]

\[\therefore \]      \[l=\int{\sqrt{\frac{1-\cos t}{1+\cos t}.}(-\,2sint.cost)dt}\]

\[=\int{\sqrt{\frac{2{{\sin }^{2}}t/2}{2{{\cos }^{2}}t/2}}(-\,2sint.cost)dt}\]

\[[\because \,\,1-cos\theta =2si{{n}^{2}}(\theta /2)\,\,and1+cos\theta =2co{{s}^{2}}(\theta /2)]\]

\[=\int{\frac{\sin t/2}{\cos t/2}\times (-\,2\times 2\times sin(t\,/2)cos(t\,/2)\times cost\,\,dt}\]

\[[\because \,sin2\theta =2sin\theta cos\theta ]\]

\[=-\int{4{{\sin }^{2}}(t\,/2)\times cos\,t\,\,dt}\]

\[=-\int{4\left( \frac{1-\cos t}{2} \right)cos\,t\,\,dt}\]           \[\left[ \because 1-\cos \theta =2si{{n}^{2}}\frac{\theta }{2} \right]\]

\[=-\,2\int{(1-cost)cos\,t\,\,dt}\]

\[=-\,2\int{(cost-co{{s}^{2}}t)dt}\]

\[=-\,2\int{\left[ \cos t-\left( \frac{1+\cos \,2t}{2} \right) \right]\,}dt\]

\[[\because \,\,cos2\theta =2co{{s}^{2}}\theta -1]\]

\[=-\,2\int{\frac{2\cos t-1-\cos 2t}{2}}\,dt\]

\[=-\int{(2\cos t-1-\cos 2t)}\,dt\]

\[=-\left( 2\sin t-t-\frac{\sin 2t}{2} \right)+C\]

\[=-\left( 2\sin t-t-\frac{2\sin t\cdot \operatorname{cost}}{2} \right)+C\]

\[=-\,2\sin t+t+\sin t.\operatorname{cost}\,+C\]

\[=-\,2\sqrt{1-{{\cos }^{2}}t}+t+\sqrt{1-{{\cos }^{2}}t}.\operatorname{cost}+C\]

\[[\because \,\,si{{n}^{2}}\theta +co{{s}^{2}}\theta =1\Rightarrow sin\theta =\sqrt{1-{{\cos }^{2}}\theta }]\]

\[=-\,2\sqrt{1-x}+{{\cos }^{-1}}\sqrt{x}+\sqrt{x}.\sqrt{1-x}+C\]

\[[\because \,\,\sqrt{x}=cost]\]