Question

$\int x^{2}Si{n}^{2}xdx=$

• A. $\frac{1}{6}{x}^3+\frac{1}{2}[\frac{x^{2}Sin2x}{2}+\frac{xCos2x}{2}-\frac{Sin2x}{4}]+c$
• B. $\frac{1}{6}{x}^3-\frac{1}{2}[\frac{x^{2}Sin2x}{2}-\frac{xCos2x}{2}+\frac{Sin2x}{4}]+C$
• C. $\frac{-1}{6}{x}^3+\frac{1}{2}[\frac{x^{2}Sin2x}{2}+\frac{xCos2x}{2}-\frac{Sin2x}{4}]+c$
• D. $\frac{1}{6}{x}^3-\frac{1}{2}[\frac{x^{2}Sin2x}{2}+\frac{xCos2x}{2}-\frac{Sin2x}{4}]+c$

Answer 1

The correct option is D $\frac{1}{6}{x}^3-\frac{1}{2}[\frac{x^{2}Sin2x}{2}+\frac{xCos2x}{2}-\frac{Sin2x}{4}]+c$

$\int x^{2}si{n}^{2}xdx$
$\int x^2(1-co{s}^{2}x)dx$
$=\int x^{2}dx-\int x^{2}co{s}^{2}xdx$
$=\frac{x^3}{3}-\int x^2\left(\frac{1+cos2x}{2}\right)dx=\frac{x^3}{3}-[\int \frac{x^2}{2}dx+\int \frac{x^{2}cos2x}{2}dx]$
$=\frac{x^3}{3}-\frac{x^3}{6}-\frac{1}{2}\int x^{2}cos2xdx.$
$\int x^{2}cos2xdx=\frac{+x^{2}sin2x}{2}-\frac{1}{2}\int 2xsin2xdx$
$\int xsin2xdx=\frac{-xcos2x}{2}+\int 1.cos2xdx$
$=\frac{-xcos2x}{2}+\frac{sin2x}{2}$
$\int x^{2}cos2xdx=\frac{+x^{2}sin2x}{2}+[\frac{-xcos2x}{2}+\frac{sin2x}{2}]$
$\int x^{2}si{n}^{2}x=\frac{x^3}{6}-\frac{1}{2}[\frac{x^{2}sin2x}{2}+\frac{xcos2x}{2}-\frac{sin2x}{4}]+c$