Integration by substitution 3e^-2t cos3tdt

Question

3e^-2t cos3tdt

Answer 1

I =  \(\int\)3e^-2tcos3t dt ..............(i)(Let)

I = 3 \(\Big[\)cos3t \(\int\)e^-2tdt - \(\int\)\(\big(\frac{d}{dt}cos3t \int e^{-2t}dt\big)dt\)\(\Big]\)

\(\Big(\)\(\because\) \(\int\)I.II dx = I\(\int\)II dx - \(\int\)\(\big(\frac{d}{dt}\)I\(\int\)II dx\(\big)\)dx\(\Big)\)(according I LATE)

= 3\(\Big[\)cos3t \(\frac{e^{-2t}}{-2}\) - \(\int\)\(\big((-)3 \sin3t\frac{e^{-2t}}{-2}\big)\)dt\(\Big]\)

= - \(\frac{3}{2}\)cos3t e^-2t - \(\frac{9}{2}\) \(\int\)sin3t e^-2t dt

= -\(\frac{3}{2}\)cos3t e^-2t - \(\frac{9}{2}\)\(\Big[\)sin3t \(\int\)e^-2t dt - \(\int \big(\frac{d}{dt}\sin3t \int e^{-2t}dt\big)dt\Big]\)

= - \(\frac{3}{2}\)cos3t e^-2t - \(\frac{9}{2}\)\(\Big[\)sin3t \(\frac{e^{-2t}}{-2}\)- \(\int\) 3cos3t \(\frac{e^{-2t}}{-2}\)dt\(\Big]\)

= - \(\frac{3}{2}\) cos3t e^-2t + \(\frac{9}{4}\)sin3t e^-2t - \(\frac{9}{4}\)\(\int\)3e^-2t cos3t dt

\(\Rightarrow\) I = - \(\frac{3}{2}\)cos3t e^-2t + \(\frac{9}{4}\)sin3t e^-2t - \(\frac{9}{4}\)I (by eqⁿ (i))

\(\Rightarrow\) I + \(\frac{9}{4}\)I = \(\frac{3}{2}\)e^-2t \(\big(\frac{3}{2}\sin3t-\cos3t)\)

\(\Rightarrow\) \(\frac{13}{4}\)I = \(\frac{3}{2}\) e^-2t \(\big(\frac{3}{2}\sin3t-\cos3t)\)

\(\Rightarrow\) 

I = \(\frac{6}{13}\)e-2t \(\big(\frac{3}{2}\sin3t-\cos3t)\)