Question

intregation of sin squar x with limit -*π÷3 to +π÷3

Answer 1

$\bold{Answer :}$

$\text{We know that,}$

$cos2x = 1 - 2 {sin}^{2}x$

$\implies{sin}^{2}x=\frac{1}{2}(1-cos2x)$

$Then, \int_{-\frac {\pi}{3}}^\frac {\pi}{3} {sin}^{2}x\:dx$

$= \frac{1}{2} \int_{-\frac {\pi}{3}}^\frac {\pi}{3} (1-cos2x)dx$

$= \frac{1}{2} \int_{-\frac{\pi}{3}}^\frac {\pi}{3} dx -\frac{1}{2} \int_{-\frac{\pi}{3}}^\frac {\pi}{3} cos2x\:dx$

$= \frac{1}{2}\bigg[x\bigg]_{-\frac{\pi}{3}}^{+\frac{\pi}{3}}-\frac{1}{4}\bigg[sin2x\bigg]_{-\frac{\pi}{3}}^{+\frac{\pi}{3}}$

$=\frac{1}{2}\{\frac{\pi}{3}-(-\frac{\pi}{3})\}-\frac{1}{4}\{sin\frac{2\pi}{3}-sin(\frac{-2\pi}{3})\}$

$=\frac{1}{2}(\frac{2\pi}{3})-\frac{1}{4}(2sin\frac{2\pi}{3})$

$=\frac{\pi}{3}-\frac{1}{4}(2\times\frac{\sqrt{3}}{2})$

$=\bold{\frac{\pi}{3}-\frac{\sqrt{3}}{4}}$

$\bigstar \: \bold{MarkAsBrainliest} \: \bigstar$