Mathematics - Definite Integration Question with Solution | TestHub

$l={\int }_0^{4\pi }{e}^t \left(si{n}^{6}2t+co{s}^4 2t\right)dt$=$l_1+l_2 {+ l}_3 {+ l}_4$
$Where l_1={\int }_0^{\pi }{e}^t \left({\sin }^{6}2t+{\cos }^{4}2t\right)dt and l_2= {\int }_{\pi }^{2\pi }{e}^t \left({\sin }^{6}2t+{cos}^4 2t\right) dt and l_3={\int }_{2\pi }^{3\pi }{e}^t\left({\sin }^{6}2t+{\cos }^{4}2t\right)dt and l_4={\int }_{3\pi }^{4\pi }{e}^t\left({sin}^{6 }2t+{cos}^4 2t\right) dt$
$t=x+\pi$
$dt=dx$
$l_2={\int }_0^{\pi }{e}^{x+\pi } \left({sin}^6 2x+{cos}^{4 }2x\right) dx= e^{\pi }{l}_1$
$t=2\pi +x$
$l_3=e^{2\pi }{l}_1$
$l_4=e^{3\pi }{l}_1$
$l=l_1 \left(1+e^{\pi }+e^{2\pi }+e^{3\pi }\right)$
$\frac{l}{l_1}=\frac{e^{4\pi }-1}{e^{\pi }-1}$
The same for a = 4.