question_answer

                          

Evaluate \[\int{\frac{\sin x+\cos x}{9+16\sin 2x}dx.}\]

OR

Evaluate \[\int{\frac{{{x}^{2}}+1}{{{(x-1)}^{2}}(x+3)}dx.}\]

 

Answer:

Let \[l=\int{\frac{\sin x+\cos x}{9+16\sin 2x}}\,dx\]

\[=\int{\frac{\sin x+\cos x}{9+16[1-{{(\sin x-\cos x)}^{2}}]}}\,dx\]

\[\left[ \begin{align}   & \because \,\,\sin 2x=2\sin x\,\cos x=2\sin x\,\cos x-1+1 \\  & =2\sin x\,\cos x-({{\cos }^{2}}x+{{\sin }^{2}}x)+1 \\  & =1-{{(\sin \,x-\cos \,x)}^{2}} \\ \end{align} \right]\]

\[=\int{\frac{(\sin x+\cos x)}{25-16{{(\sin x-\cos x)}^{2}}}\,dx}\]

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

\[\Rightarrow \]   \[dt=(\cos x+\sin x)\,dt\]

\[\therefore \]      \[l=\int{\frac{dt}{25-16{{t}^{2}}}=\frac{1}{16}}\int{\frac{dt}{{{\left( \frac{5}{4} \right)}^{2}}-{{t}^{2}}}}\]

\[=\frac{1}{16}\cdot \frac{1}{2}\cdot \frac{4}{5}\log \left| \frac{\frac{5}{4}+t}{\frac{5}{4}-t} \right|+C\]

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

\[=\frac{1}{40}\log \left| \frac{5+4(\sin x-\cos x)}{5-4(\sin x-\cos x)} \right|+C\]

\[[\text{put}\,\,t=\sin x-\cos x]\]

OR

Let \[l=\int{\frac{{{x}^{2}}+1}{{{(x+1)}^{2}}(x+3)}dx}\]

Again, let \[\frac{{{x}^{2}}+1}{{{(x+1)}^{2}}(x+3)}=\frac{A}{x-1}+\frac{B}{{{(x-1)}^{2}}}+\frac{C}{x+3}\] ?(i)

\[\Rightarrow \] \[{{x}^{2}}+1=A\,(x-1)\,(x+3)+B\,(x+3)+C\,{{(x-1)}^{2}}\]? (ii)

On putting x = 1 in Eq. (ii), we get

2 = 4B  \[\Rightarrow \] \[B=\frac{1}{2}\]

On putting \[x=-\,3\] in Eq. (ii), we get

10 = 16C \[\Rightarrow \] \[C=\frac{10}{16}=\frac{5}{8}\]

On equating coefficient of \[{{x}^{2}}\] from both sides of Eq. (ii), we get

\[1=A+C\]\[\Rightarrow \] \[A=1-C=1-\frac{5}{8}=\frac{3}{8}\]

Now, on substituting the values of A, B and C in

Eq. (i), we get

\[\frac{{{x}^{2}}+1}{{{(x-1)}^{2}}(x+3)}=\frac{3}{8}\cdot \frac{1}{(x-1)}+\frac{1}{2}\cdot \frac{1}{{{(x-1)}^{2}}}+\frac{5}{8}\cdot \frac{1}{(x+3)}\]On integrating both sides, we get

\[\int{\frac{{{x}^{2}}+1}{{{(x-1)}^{2}}(x+3)}\,dx=\int{\left[ \frac{3}{8}\frac{1}{(x-1)}+\frac{1}{2}\frac{1}{{{(x-1)}^{2}}}+\frac{5}{8}\frac{1}{(x+3)} \right]}\,dx}\]\[\therefore \,\,l=\frac{3}{8}\int{\frac{1}{(x-1)}\,dx}+\frac{1}{2}\int{\frac{1}{{{(x-1)}^{2}}}\,dx}+\frac{5}{8}\int{\frac{1}{(x+3)}\,dx}\]\[=\frac{3}{8}\,\log \,\,|x-1|-\frac{1}{2(x-1)}+\frac{5}{8}\,\log \,\,|x+3|+\,C\]