Answer
$$\int\sin t\sqrt{1+\cos t}dt=-\frac{2}{3}\sqrt{(1+\cos t)^3}+C$$
Work Step by Step
\int$$A=\int\sin t\sqrt{1+\cos t}dt$$
Let $u=1+\cos t$
Then $du=(1+\cos t)'dt=-\sin tdt$. So $\sin tdt=-du$
Also, $\sqrt{1+\cos t}=\sqrt u=u^{1/2}$
Substitute into $A$, we have $$A=-\int u^{1/2}du$$ $$A=-\frac{u^{3/2}}{\frac{3}{2}}+C$$ $$A=-\frac{2\sqrt{u^3}}{3}+C$$ $$A=-\frac{2}{3}\sqrt{(1+\cos t)^3}+C$$
