## Calculus: Early Transcendentals 8th Edition

$$\int^{\pi/2}_0\cos^2\theta d\theta=\frac{\pi}{4}$$
$$A=\int^{\pi/2}_0\cos^2\theta d\theta$$ Here we see that the power of $\cos$ is even, so if we change $\cos^2\theta=1-\sin^2\theta$, we still do not have enough elements of both $\cos$ and $\sin$ to use the Substitution Rule. However, in this case, we have $$\cos2\theta=2\cos^2 \theta-1$$ That means $$\cos^2 \theta=\frac{1}{2}(\cos 2\theta+1)$$ Therefore, $$A=\frac{1}{2}\int^{\pi/2}_0(\cos 2\theta+1)d\theta$$ $$A=\frac{1}{2}\Bigg[\int^{\pi/2}_0\cos2\theta d\theta+\int^{\pi/2}_01d\theta\Bigg]$$ $$A=\frac{1}{2}\Bigg[\frac{1}{2}\int^{\pi/2}_0\cos2\theta d(2\theta)+\int^{\pi/2}_01d\theta\Bigg]$$ $$A=\frac{1}{2}\Bigg[\frac{1}{2}\sin2\theta+\theta\Bigg]^{\pi/2}_0$$ $$A=\Bigg[\frac{1}{4}\sin2\theta+\frac{1}{2}\theta\Bigg]^{\pi/2}_0$$ $$A=\Bigg[\frac{1}{4}\sin2\pi+\frac{1}{2}\frac{\pi}{2}\Bigg]-\Bigg[\frac{1}{4}\sin0+\frac{1}{2}\times0\Bigg]$$ $$A=\Bigg[\frac{1}{4}\times0+\frac{\pi}{4}\Bigg]-0$$ $$A=\frac{\pi}{4}$$