evaluate the following integral.\n int_{0}^{pi} 4sin^{4}x dx \n int_{0}^{pi} 4sin^{4}x dx=square (type an…

Explanation:

Step1: Use power - reduction formula

We know that $\sin^{2}x=\frac{1 - \cos(2x)}{2}$. So, $\sin^{4}x = (\sin^{2}x)^{2}=\left(\frac{1 - \cos(2x)}{2}\right)^{2}=\frac{1 - 2\cos(2x)+\cos^{2}(2x)}{4}$. Then the integral $\int_{0}^{\pi}4\sin^{4}x dx=\int_{0}^{\pi}(1 - 2\cos(2x)+\cos^{2}(2x))dx$.

Step2: Use power - reduction formula for $\cos^{2}(2x)$

Since $\cos^{2}(u)=\frac{1+\cos(2u)}{2}$, for $u = 2x$, $\cos^{2}(2x)=\frac{1+\cos(4x)}{2}$. The integral becomes $\int_{0}^{\pi}\left(1 - 2\cos(2x)+\frac{1+\cos(4x)}{2}\right)dx=\int_{0}^{\pi}\left(\frac{2 - 4\cos(2x)+1+\cos(4x)}{2}\right)dx=\int_{0}^{\pi}\left(\frac{3 - 4\cos(2x)+\cos(4x)}{2}\right)dx$.

Step3: Integrate term - by - term

$\int_{0}^{\pi}\frac{3}{2}dx-\int_{0}^{\pi}2\cos(2x)dx+\int_{0}^{\pi}\frac{1}{2}\cos(4x)dx$. For $\int_{0}^{\pi}\frac{3}{2}dx=\frac{3}{2}x\big|{0}^{\pi}=\frac{3\pi}{2}$. For $\int{0}^{\pi}2\cos(2x)dx$, let $u = 2x$, $du=2dx$. When $x = 0$, $u = 0$; when $x=\pi$, $u = 2\pi$. So $\int_{0}^{\pi}2\cos(2x)dx=\sin(2x)\big|{0}^{\pi}=0$. For $\int{0}^{\pi}\frac{1}{2}\cos(4x)dx$, let $v = 4x$, $dv = 4dx$. When $x = 0$, $v = 0$; when $x=\pi$, $v = 4\pi$. So $\int_{0}^{\pi}\frac{1}{2}\cos(4x)dx=\frac{1}{8}\sin(4x)\big|_{0}^{\pi}=0$.

Answer:

$\frac{3\pi}{2}$