2
$\begingroup$

I need to solve this integral: $$ \int{x\sin^2(x)}\ dx $$

I SOLVED it by writting: $$ \sin^2(x) = \frac{1-\cos(2x)}{2} $$ and used integration by parts for $x\cos(2x)$, and the result is:

$$ \frac{1}{4}(x^2-x\sin(2x)-\frac{\cos(2x)}{2}) + C $$

I need to know other methods of solving this integral (preferable simple methods). Thank you!

$\endgroup$
  • $\begingroup$ $ \int x \sin^2(x) dx = \frac{1}{2} \int (x-x\cos(2x)) dx $ Use integration by parts on the $x\cos(2x)$. $\endgroup$ – bigfocalchord Jul 19 '16 at 12:23
  • $\begingroup$ Integration by parts. $\endgroup$ – MathsPro Jul 19 '16 at 12:23
  • $\begingroup$ @dydxx I used that method, I want to know other methods. Check my question. $\endgroup$ – MM PP Jul 19 '16 at 12:24
  • $\begingroup$ Do you specifically want to avoid integration by parts? $\endgroup$ – user 170039 Jul 19 '16 at 12:28
  • $\begingroup$ @user170039 No. I just want different methods of solving this. I used the method I added in the question, but I want to know as many methods as possible. Thank you. $\endgroup$ – MM PP Jul 19 '16 at 12:29
6
$\begingroup$

Hint:- Let $$I=\displaystyle\int x\sin^2 x\ dx$$$$J=\displaystyle\int x\cos^2 x\ dx$$Now observe that, $$I+J=\displaystyle\int x\ dx$$and $$I-J=\Re\left(\displaystyle\int xe^{2ix}\ dx\right)$$where $\Re\left(\displaystyle\int xe^{2ix}\ dx\right)$ denotes the real part of the integral.

$\endgroup$
1
$\begingroup$

Although this is a more complicated sotution way ,but different from your method $$\int { x\sin ^{ 2 } (x) } dx=\int { \sin ^{ 2 } (x)d\left( \frac { { x }^{ 2 } }{ 2 } \right) =\frac { { x }^{ 2 }\sin ^{ 2 } (x) }{ 2 } } -\frac { 1 }{ 2 } \int { { x }^{ 2 }\sin { 2x } dx } =\\ =\frac { { x }^{ 2 }\sin ^{ 2 } (x) }{ 2 } +\frac { 1 }{ 4 } \int { { x }^{ 2 }d\cos { 2x } =\frac { { x }^{ 2 }\sin ^{ 2 } (x) }{ 2 } +\frac { 1 }{ 4 } \left( { x }^{ 2 }\cos { 2x } -2\int { x\cos { 2xdx } } \right) = } \\ =\frac { { x }^{ 2 }\sin ^{ 2 } (x) }{ 2 } +\frac { 1 }{ 4 } \left( { x }^{ 2 }\cos { 2x } -\int { xd\sin { 2x } } \right) =\frac { { x }^{ 2 }\sin ^{ 2 } (x) }{ 2 } +\frac { 1 }{ 4 } \left( { x }^{ 2 }\cos { 2x } -x\sin { 2x+\int { \sin { 2xdx } } } \right) =\\ =\frac { { x }^{ 2 }\sin ^{ 2 } (x) }{ 2 } +\frac { { x }^{ 2 }\cos { 2x- } x\sin { 2x } }{ 4 } -\frac { 1 }{ 8 } \cos { 2x } +C$$

$\endgroup$
1
$\begingroup$

$$f(x)=\int x\sin(x)^2dx$$ $$f(x)=-\frac 14\int x\left(e^{2ix}+e^{-2ix}-2\right)dx$$

By applying the Tabular Method for Integration by parts you get

$$f(x)=-\frac 14\left(\frac x{2i}e^{2ix}+\frac 14 e^{2ix}-\frac x{2i}e^{-2ix}+\frac 14e^{-2ix}-x^2\right)+C$$

Gathering the terms, yields the same result.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.