How do you integrate $\int x sin 2 x$ $int xsin2x$ by parts from $[0, π]$ $[0,pi]$ ?

Question

How do you integrate $\int x sin 2 x$ $int xsin2x$ by parts from $[0, π]$ $[0,pi]$ ?

1 Answer

Impact of this question

3726 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-05T13:59:58

$\int_{0}^{π} x sin 2 x d x = - \frac{π}{2}$ $int_0^pi xsin2x dx =-pi/2$

Explanation:

Consider the integral:

$\int_{0}^{π} x sin 2 x d x$ $int_0^pi xsin2x dx$

Note that:

$d (- \frac{1}{2} cos 2 x) = sin 2 x d x$ $d(-1/2cos2x) = sin2x dx$

so we can integrate by parts:

$\int_{0}^{π} x sin 2 x d x = \int_{0}^{π} x d (- \frac{1}{2} cos 2 x)$ $int_0^pi xsin2x dx = int_0^pi xd(-1/2cos2x)$

$\int_{0}^{π} x sin 2 x d x = {[- \frac{x cos 2 x}{2}]}_{0}^{π} + \frac{1}{2} \int_{0}^{π} cos 2 x d x$ $int_0^pi xsin2x dx = [-(xcos2x)/2]_0^pi +1/2 int _0^pi cos2x dx$

$\int_{0}^{π} x sin 2 x d x = {[- \frac{x cos 2 x}{2}]}_{0}^{π} + \frac{1}{4} {[sin 2 x]}_{0}^{π}$ $int_0^pi xsin2x dx = [-(xcos2x)/2]_0^pi +1/4 [sin2x]_0^pi$

$\int_{0}^{π} x sin 2 x d x = [- \frac{π cos (2 π)}{2} + 0] + [0 - 0]$ $int_0^pi xsin2x dx = [-(picos(2pi))/2 +0] + [0 -0]$

$\int_{0}^{π} x sin 2 x d x = - \frac{π}{2}$ $int_0^pi xsin2x dx =-pi/2$

Science

Math

Humanities

... and beyond

How do you integrate $\int x sin 2 x$ $int xsin2x$ by parts from $[0, π]$ $[0,pi]$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you integrate int xsin2x∫xsin2xint xsin2x by parts from [0,pi][0,π][0,pi]?

1 Answer

Explanation:

Related questions

Impact of this question

How do you integrate $\int x sin 2 x$ $int xsin2x$ by parts from $[0, π]$ $[0,pi]$ ?