University Calculus: Early Transcendentals (3rd Edition)

$\dfrac{3 \pi}{8}+1$
Re-arrange the integral as follows: $\int_{0}^{\frac{\pi}{2}} \dfrac{(1+\sin \phi)^2}{2} d \phi=\int_{0}^{\frac{\pi}{2}} \dfrac{(1+\sin^2 \phi+2 \sin \phi)}{2} d \phi$ This implies that $\int_{0}^{\frac{\pi}{2}} \dfrac{(1+\sin^2 \phi+2 \sin \phi)}{2} d \phi= (\dfrac{3}{4}+\sin\phi-\dfrac{\cos 2 \phi}{4}) d \phi|_{0}^{\frac{\pi}{2}}-\sin 2 \phi|_{0}^{\frac{\pi}{2}}$ Thus, we have $(\dfrac{3}{4}+\sin\phi-\dfrac{\cos 2 \phi}{4}) d \phi|_{0}^{\frac{\pi}{2}}-\sin 2 \phi|_{0}^{\frac{\pi}{2}}=\dfrac{3 \pi}{8}+1$