trigonometric integral: int [x cos^2(x)] dx

[sleepaholic]

the integral of x cos^2(x) dx

 

[royhaas]

Write \(\displaystyle cos^2(x)\) as a function of \(\displaystyle cos(2x)\), then use integration by parts.

 

[galactus]

As Roy suggested:

I will start you off:

\(\displaystyle \L\\\int{xcos^{2}(x)}dx\)

Sub in \(\displaystyle cos^{2}(x)=\frac{1+cos(2x)}{2}\)

\(\displaystyle \L\\\int{\frac{1}{2}x+\frac{1}{2}cos(2x)}dx\)

\(\displaystyle \L\\\frac{1}{2}\int{x}dx+\int{xcos(2x)dx\)

\(\displaystyle \L\\\frac{1}{4}x^{2}+\frac{1}{2}\int{xcos(2x)}dx\)

Now, let \(\displaystyle u=2x;\;\ \frac{du}{2}=dx\)

 

[sleepaholic]

but if i do u substitution then there is an x left over or am i supposed to say that x=u/2?
(1/4)x^2 + (1/4) int (u/2) cosu du
how do i get take the antideriv when there is a variable infront? am i supposed to use u dv stuff?

 

[galactus]

Yes, go ahead and sub. Since u=2x and x=u/2

\(\displaystyle \L\\\frac{1}{4}x^{2}+\frac{1}{2}\int(\frac{1}{2}ucos(u))du\)

\(\displaystyle \L\\\frac{1}{4}x^{2}+\frac{1}{4}\int\underbrace{ucos(u)du}_{\text{integration by parts}}\)

 

[sleepaholic]

i'm still confused...bc i tried doing integration by parts and i keep going in circles with it.

 

[stapel]

[I'm still confused[, because I] tried doing integration by parts and keep going in circles with it.

Please reply showing your steps. Thank you.

Eliz.

 

[galactus]

Let's just start from scratch. Forget about the u-subbing in my last post.

\(\displaystyle \L\\\int{xcos^{2}(x)}dx\)

Let \(\displaystyle \L\\cos^{2}(x)=\frac{1}{2}+\frac{cos(2x)}{2}\)

\(\displaystyle \L\\\int{x(\frac{1}{2}+\frac{cos(2x)}{2})dx\)

\(\displaystyle \L\\\int\frac{1}{2}xdx+\int\frac{1}{2}xcos(2x)dx\)

\(\displaystyle \L\\\frac{1}{2}\int{x}dx+\frac{1}{2}\int{xcos(2x)}dx\)

The first half you know. That's easy.

Let \(\displaystyle u=x; \;\ dv=cos(2x)dx; \;\ du=dx; \;\ v=\frac{sin(2x)}{2}\)