$$\int_{0}^{\pi} e^t\left(\sin^6{at} +\cos^4{at}\right)dt $$
$$\int e^t\left(1-2\sin^2{at}\cos^2{at} - \sin^4{at}\cos^2{at}\right) $$ $$\int e^t\left(1-\sin^2{at}\cos^2{at}\left(2 +\frac{1-\cos{2at}}{2}\right) \right)$$ $$ \int e^t -\frac58e^t\sin^2{2at} +\frac18 e^t\sin^2{2at}\cos{2at} $$ $$=e^{\pi}-1$$ It can be further solved by Integrating by part. But this is competition question so it is becoming lengthy. Is there any other way to solve it quickly?
HINT: the trigonometric sum in the brackets can be simplified to $$\frac{1}{32} (\cos (2 a t)+10 \cos (4 a t)-\cos (6 a t)+22)$$