question_answer

Evaluate \[\int{\sqrt{3-4x-4{{x}^{2}}}\,\,dx.}\]

OR

Evaluate \[\int{\frac{\sin (x-\alpha )}{\sin (x+\alpha )}\,dx.}\]

Answer:

Let \[l=\int{\sqrt{3-4x-4{{x}^{2}}}\,\,dx=\int{\sqrt{-\,4\left( {{x}^{2}}+x-\frac{3}{4} \right)}}\,dx}\]

\[=\sqrt{4}\int{\sqrt{-\left( {{x}^{2}}+x-\frac{3}{4} \right)}}\,dx\]

\[=2\int{\sqrt{-\left( {{x}^{2}}+x-\frac{3}{4}+\frac{1}{4}-\frac{1}{4} \right)}}\,dx\]

\[\left[ \text{adding and subtracting}\,\,\frac{1}{4} \right]\]

\[=\sqrt[2]{\left[ -{{\left( x+\frac{1}{2} \right)}^{2}}+\left( \frac{3}{4}+\frac{1}{4} \right) \right]}\,dx\]

\[\left[ \because \,\,{{x}^{2}}+x+\frac{1}{4}={{\left( x+\frac{1}{2} \right)}^{2}} \right]\]

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

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

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

\[=2\left[ \frac{2x+1}{4}\sqrt{\frac{3-4x-4{{x}^{2}}}{4}}+\frac{1}{2}{{\sin }^{-1}}\left( \frac{2x+1}{2} \right) \right]+C\]\[\left[ \because \,\,\sqrt{1-{{\left( x+\frac{1}{2} \right)}^{2}}}=\sqrt{1-{{x}^{2}}-x-\frac{1}{4}}=\sqrt{\frac{3-4{{x}^{2}}-4x}{4}} \right]\]\[=\frac{2x+1}{4}\sqrt{3-4x-4{{x}^{2}}}+{{\sin }^{-1}}\left( \frac{2x+1}{2} \right)+C\]

OR

Let \[l=\,\int{\frac{\sin (x-\alpha )}{\sin (x+\alpha )}\,dx.}\]

Put \[x+\alpha =t\] \[\Rightarrow \] \[x=t-\alpha \] \[\Rightarrow \] dx = dt

Then, given integral reduces to \[l=\,\int{\frac{\sin (t-2\alpha )}{\sin \,t}\,dt}\]

\[\Rightarrow \]   \[l=\int{\frac{\sin t\cos 2\alpha -\cos t\sin 2\alpha }{\sin t}dt}\]

\[[\because \sin (x-y)=\sin x\cos y-\cos x\sin y]\]

\[\Rightarrow \]   \[l=\int{\left( \frac{\sin t\cos 2\alpha }{\sin t}-\frac{\cos t\sin 2\alpha }{\sin t} \right)\,dt}\]

\[\left[ \because \,\,\frac{\cos x}{\sin x}=\cot \,x \right]\]

\[\Rightarrow \]   \[l=\cos 2\alpha \int{dt-\sin 2\alpha \int{\cot t\,dt}}\]

\[=t\cos 2\alpha -\sin 2\alpha \cdot \log |\sin t|+C\]

\[[\because \,\,\int{\cot \,x\,dx=\log |\sin \,x|]}\]

\[\therefore \] \[l=(x+\alpha )\cos 2\alpha -\sin 2\alpha \cdot log|\sin (x+\alpha )|+\,C\]

\[[\text{put}\,\,t=x+\alpha ]\]