question_answer

Evaluate \[\int{\frac{dx}{\sin (x-a)\cdot \cos (x-b)}}.\]

OR

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

Answer:

Let \[I=\int{\frac{dx}{\sin (x-a)\cdot \cos (x-b)}}\]

\[=\frac{1}{\cos (b-a)}\int{\frac{\cos [(x-a)-(x-b)]}{\sin (x-a)\cdot \cos (x-b)}\,}dx\]

[Multiply numerator and denominator by\[cos(b-a)\]]

\[=\frac{1}{\cos (b-a)}\int{\frac{\left[ \begin{align}   & \cos (x-a)(x-b) \\  & +\sin (x-a)\sin (x-b) \\ \end{align} \right]}{\sin (x-a)\cdot \cos (x-b)}}\,dx\]

\[[\because \cos (A-B)=\cos A\cos B+\sin A\sin B]\]

\[=\frac{1}{\cos (b-a)}\left\{ \int{\left[ \frac{\cos (x-a)\cos (x-b)}{\sin (x-a)\cos (x-b)} \right]} \right.dx\]

\[+\left. \int{\left[ \frac{\sin (x-a)\sin (x-b)}{\sin (x-a)\cos (x-b)} \right]dx} \right\}\]

\[=\frac{1}{\cos (b-a)}[\int{\cot (x-a)dx+\int{\tan (x-b)dx}}]\]

\[=\frac{1}{\cos (b-a)}[\log |\sin (x-a)|-\log |\cos (x-b)|+\,C]\]\[\mathbf{Z=4x+3y}\]                                              \[\int{\tan x\,dx=-\log |\cos \,x|+\,C}]\]

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

\[\left[ \because \,\,\,\log \,\,m-\log \,\,n=\log \frac{m}{n} \right]\]

OR

Let \[I=\int{\frac{x{{e}^{2x}}}{{{(1+2x)}^{2}}}}\,dx\]

Put \[2x=t\,\,\,\Rightarrow \,\,\,2=\frac{dt}{dx}\,\,\,\Rightarrow \,\,\,dx=\frac{dt}{2}\]

\[\therefore \]      \[I=\int{\frac{\frac{t}{2}{{e}^{t}}}{{{(1+t)}^{2}}}}\,\frac{dt}{2}=\frac{1}{4}\int{\frac{t{{e}^{t}}}{{{(1+t)}^{2}}}}\,dt\]

\[=\frac{1}{4}\int{\frac{(1+t-1){{e}^{t}}}{{{(1+t)}^{2}}}}\,dt\]

\[=\frac{1}{4}\int{{{e}^{t}}\left[ \frac{(1+t)}{{{(1+t)}^{2}}}-\frac{1}{{{(1+t)}^{2}}} \right]}\,dt\]

\[=\frac{1}{4}\int{{{e}^{t}}\left[ \frac{1}{(1+t)}-\frac{1}{{{(1+t)}^{2}}} \right]}\,dt\]

On comparing with

\[\int{{{e}^{t}}[f(t)+f'(t)]dt={{e}^{t}}f(t)+C,}\] we get

\[f(t)=\frac{1}{(1+t)},\] \[f'(t)=\frac{-\,1}{{{(1+t)}^{2}}}\]

\[\therefore \]      \[I=\frac{1}{4}{{e}^{t}}.\frac{1}{1+t}+C\]

\[=\frac{{{e}^{2x}}}{4(1+2x)}+C\]                 [put\[t=2x\]]