Question

rewrite the given integral using this substitution.
int_{1}^{2}sqrt{4 - x^{2}} dx=int_{\frac{pi}{6}}^{square}(square) d\theta
(type exact answers.)

Explanation:

Step1: Use substitution $x = 2\sin\theta$

If $x = 2\sin\theta$, then $dx=2\cos\theta d\theta$. Also, $\sqrt{4 - x^{2}}=\sqrt{4 - 4\sin^{2}\theta}=2\cos\theta$. When $x = 1$, $1 = 2\sin\theta$, so $\sin\theta=\frac{1}{2}$ and $\theta=\frac{\pi}{6}$. When $x = 2$, $2 = 2\sin\theta$, so $\sin\theta = 1$ and $\theta=\frac{\pi}{2}$.

Step2: Rewrite the integral

The integral $\int_{1}^{2}\sqrt{4 - x^{2}}dx$ becomes $\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}(2\cos\theta)\times(2\cos\theta)d\theta=\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}4\cos^{2}\theta d\theta$.

Answer:

$\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}4\cos^{2}\theta d\theta$