Find the exact value of $\int_{0}^{\frac{1}{2} π} x^{2} sin 2 x d x$ $int_0^(1/2pi)x^2sin2x dx$ ?

Question

Find the exact value of $\int_{0}^{\frac{1}{2} π} x^{2} sin 2 x d x$ $int_0^(1/2pi)x^2sin2x dx$ ?

1 Answer

Impact of this question

1684 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-17T18:59:17

$\int_{0}^{\frac{π}{2}} x^{2} \cdot sin 2 x d x = \frac{π^{2}}{8} - \frac{1}{2}$ $int_0^(pi/2) x^2*sin2xdx=pi^2/8-1/2$

Explanation:

Here,

$I = \int_{0}^{\frac{π}{2}} x^{2} \cdot sin 2 x d x$ $I=int_0^(pi/2) x^2*sin2xdx$

Using Integration by parts:

$I = {[x^{2} \cdot (- \frac{cos (2 x)}{2})]}_{0}^{2} - \int_{0}^{\frac{π}{2}} 2 x (- \frac{cos (2 x)}{2}) d x$ $I=[x^2*(-cos(2x)/2)]_0^(pi/2) -int_0^(pi/2)2x(-cos(2x)/2)dx$

$= [\frac{π^{2}}{4} (- \frac{cos π}{2}) - 0] + \int_{0}^{\frac{π}{2}} x cos 2 x d x$ $=[pi^2/4(-cospi/2)-0]+int_0^(pi/2)xcos2xdx$

Again using Integration by parts:

$= [\frac{π^{2}}{4} (- \frac{- 1}{2})] + {[x (\frac{sin (2 x)}{2})]}_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} 1 (\frac{sin (2 x)}{2}) d x$ $=[pi^2/4(-(-1)/2)]+[x(sin(2x)/2)]_0^(pi/2) - int_0^(pi/2)1(sin(2x)/2)dx$

$= \frac{π^{2}}{8} + [\frac{π}{2} \cdot \frac{sin π}{2} - 0] - \frac{1}{2} {[- \frac{cos (2 x)}{2}]}_{0}^{\frac{π}{2}}$ $=pi^2/8+[pi/2*sinpi/2-0]-1/2[-cos(2x)/2]_0^(pi/2)$

$= \frac{π^{2}}{8} + 0 + \frac{1}{4} [cos π - cos 0]$ $=pi^2/8+0+1/4[cospi-cos0]$

$= \frac{π^{2}}{8} + \frac{1}{4} [- 1 - 1]$ $=pi^2/8+1/4[-1-1]$

$:.I=pi^2/8-1/2$

Science

Math

Humanities

... and beyond

Find the exact value of $\int_{0}^{\frac{1}{2} π} x^{2} sin 2 x d x$ $int_0^(1/2pi)x^2sin2x dx$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

Find the exact value of int_0^(1/2pi)x^2sin2x dx∫12π0x2sin2xdxint_0^(1/2pi)x^2sin2x dx?

1 Answer

Explanation:

Related questions

Impact of this question

Find the exact value of $\int_{0}^{\frac{1}{2} π} x^{2} sin 2 x d x$ $int_0^(1/2pi)x^2sin2x dx$ ?