Answer
$$\int^{\pi}_05(5-4\cos t)^{1/4}\sin tdt=3^{5/2}-1$$
Work Step by Step
$$A=\int^{\pi}_05(5-4\cos t)^{1/4}\sin tdt$$
We set $u=5-4\cos t$, which means $$du=-4(-\sin t)dt=4\sin tdt$$ $$\sin tdt=\frac{1}{4}dt$$
For $t=\pi$, we have $u=5-4\cos\pi=5-4(-1)=9$
For $t=0$, we have $u=5-4\cos0=5-4(1)=1$
Therefore, $$A=\frac{5}{4}\int^9_1u^{1/4}du=\frac{5}{4}\times\frac{u^{5/4}}{\frac{5}{4}}\Big]^9_1=u^{5/4}\Big]^9_1$$ $$A=9^{5/4}-1$$ $$A=(3^2)^{5/4}-1$$ $$A=3^{5/2}-1$$
