Question

integral 0 to pi bi 2 sin raise to 4 ​

Answer 1

We have, if $n\in\mathbb{N},$

$\displaystyle\small\text{ \longrightarrow\int\limits_0^{\frac{\pi}{2}}\sin^nx\ dx=\int\limits_0^{\frac{\pi}{2}}\cos^nx\ dx=\left\{\begin{array}{ll}\dfrac{(n-1)(n-3)(n-5)\dots2}{n(n-2)(n-4)\dots3},&n\ is\ odd\\\\\dfrac{(n-1)(n-3)(n-5)\dots3}{n(n-2)(n-4)\dots2}\cdot\dfrac{\pi}{2},&n\ is\ even\end{array}\right. }$

Here, we've to evaluate,

$\displaystyle\longrightarrow I=\int\limits_0^{\frac{\pi}{2}}\sin^4x\ dx$

Here $n=4$ is even. Hence,

$\longrightarrow I=\dfrac{(n-1)(n-3)(n-5)\dots3}{n(n-2)(n-4)\dots2}\cdot\dfrac{\pi}{2}$

$\longrightarrow I=\dfrac{3}{4\cdot2}\cdot\dfrac{\pi}{2}$

$\longrightarrow\underline{\underline{I=\dfrac{3\pi}{16}}}$

Hence 3π/16 is the answer.