question\nuse substitution to evaluate the definite integral given below.\n int_{sqrt{\frac{pi}{3}}}^{sqrt{\f…

Explanation:

Step1: Set the substitution

Let $u = 3x^{2}$. Then $du=6x dx$, and $x dx=\frac{1}{6}du$.

Step2: Find new limits of integration

When $x = \sqrt{\frac{\pi}{3}}$, $u = 3\times(\sqrt{\frac{\pi}{3}})^{2}=\pi$. When $x=\sqrt{\frac{\pi}{12}}$, $u = 3\times(\sqrt{\frac{\pi}{12}})^{2}=\frac{\pi}{4}$.

Step3: Rewrite the integral

The integral $\int_{\sqrt{\frac{\pi}{3}}}^{\sqrt{\frac{\pi}{12}}}x\sin(3x^{2})\cos^{4}(3x^{2})dx$ becomes $\frac{1}{6}\int_{\pi}^{\frac{\pi}{4}}\sin(u)\cos^{4}(u)du$.

Step4: Set a new - substitution for simplicity

Let $t=\cos(u)$, then $dt =-\sin(u)du$. When $u=\pi$, $t=- 1$; when $u = \frac{\pi}{4}$, $t=\frac{\sqrt{2}}{2}$. The integral $\frac{1}{6}\int_{\pi}^{\frac{\pi}{4}}\sin(u)\cos^{4}(u)du$ can be rewritten as $-\frac{1}{6}\int_{-1}^{\frac{\sqrt{2}}{2}}t^{4}dt$.

Step5: Integrate the power - function

The antiderivative of $t^{4}$ is $\frac{t^{5}}{5}$. So, $-\frac{1}{6}\int_{-1}^{\frac{\sqrt{2}}{2}}t^{4}dt=-\frac{1}{6}\left[\frac{t^{5}}{5}\right]_{-1}^{\frac{\sqrt{2}}{2}}$.

Step6: Evaluate the definite integral

$-\frac{1}{30}\left[\left(\frac{\sqrt{2}}{2}\right)^{5}-(-1)^{5}\right]=-\frac{1}{30}\left[\frac{\sqrt{2}}{8}+ 1\right]=-\frac{\sqrt{2}+8}{240}$.

Answer:

$-\frac{\sqrt{2}+8}{240}$