Question

evaluate $int_{0}^{pi/8}10sqrt{1 - cos8x}dx$.
$int_{0}^{pi/8}10sqrt{1 - cos8x}dx=square$ (simplify your answer, including any radicals. use integers or fractions)

Explanation:

Step1: Use the half - angle identity

Recall the identity $1-\cos(2\theta)=2\sin^{2}\theta$. Let $\theta = 4x$, then $1 - \cos(8x)=2\sin^{2}(4x)$. So the integral becomes $\int_{0}^{\frac{\pi}{8}}10\sqrt{2\sin^{2}(4x)}dx$.

Step2: Simplify the integrand

Since for $x\in[0,\frac{\pi}{8}]$, $\sin(4x)\geq0$, $\sqrt{2\sin^{2}(4x)}=\sqrt{2}\sin(4x)$. The integral is now $\int_{0}^{\frac{\pi}{8}}10\sqrt{2}\sin(4x)dx$.

Step3: Use substitution

Let $u = 4x$, then $du=4dx$. When $x = 0$, $u = 0$; when $x=\frac{\pi}{8}$, $u=\frac{\pi}{2}$. The integral becomes $\frac{10\sqrt{2}}{4}\int_{0}^{\frac{\pi}{2}}\sin(u)du$.

Step4: Evaluate the integral

We know that $\int\sin(u)du=-\cos(u)+C$. So $\frac{10\sqrt{2}}{4}[-\cos(u)]_{0}^{\frac{\pi}{2}}$.

Step5: Substitute the limits of integration

$\frac{10\sqrt{2}}{4}[-\cos(\frac{\pi}{2})+\cos(0)]=\frac{10\sqrt{2}}{4}(0 + 1)=\frac{5\sqrt{2}}{2}$.

Answer:

$\frac{5\sqrt{2}}{2}$