## University Calculus: Early Transcendentals (3rd Edition)

$$\int e^\theta\sin\theta d\theta=\frac{e^\theta}{2}(\sin\theta-\cos\theta)+C$$
$$A=\int e^\theta\sin\theta d\theta$$ Take $u=\sin\theta$ and $dv=e^\theta d\theta$ We then have $du=\cos\theta d\theta$ and $v=e^\theta$ Apply the formula $\int udv= uv-\int vdu$, we have $$A=e^\theta\sin\theta-\int e^\theta\cos\theta d\theta$$ We carry out integration by parts one more time: Take $u=\cos\theta$ and $dv=e^\theta d\theta$ We then have $du=-\sin\theta d\theta$ and $v=e^\theta$ Apply the formula $\int udv= uv-\int vdu$, we have $$A=e^\theta\sin\theta-\Big(e^\theta\cos\theta-\int -e^\theta\sin\theta d\theta\Big)$$ $$A=e^\theta\sin\theta-\Big(e^\theta\cos\theta+\int e^{\theta}\sin\theta d\theta\Big)$$ $$A=e^\theta\sin\theta-e^\theta\cos\theta-\int e^{\theta}\sin\theta d\theta$$ We notice here that $\int e^{\theta}\sin\theta d\theta$ is in fact the given integral, which means it equals $A$. Therefore, $$A=e^\theta\sin\theta-e^\theta\cos\theta-A$$ $$2A=e^\theta\sin\theta-e^\theta\cos\theta+C$$ $$A=\frac{e^\theta\sin\theta-e^\theta\cos\theta}{2}+C=\frac{e^\theta}{2}(\sin\theta-\cos\theta)+C$$