Question

∫x(sinx)^2 dx

Answer 1

Hmm so parts and stuff huh? :)

Answer 2

Remember your `Sine Half-Angle Identity`?

Answer 3

yes its (1+cos2x)/2

Answer 4

\[\Large\bf\sf \sin^2x\quad=\quad \frac{1}{2}(1-\cos2x)\]Minus inside for sine. But yes close.

Answer 5

oh right I gave the one for cosine

Answer 6

\[\Large\bf\sf \frac{1}{2}\int\limits x-x \cos2x\; dx\]So we just need to apply int-by-parts to the second term I guess.

Answer 7

\[\Large\bf\sf \frac{1}{4}x^2-\int\limits x \cos 2x\;dx\]Might help to get that first term out of the way.
Understand how to setup your `parts`?

Answer 8

why did the x come out as x^2 ?

Answer 9

I guess I jumped too far ahead :x my bad.\[\Large\bf\sf \frac{1}{2}\int\limits\limits x-x \cos2x\; dx\quad=\quad \frac{1}{2}\int\limits x\;dx-\frac{1}{2}\int\limits x \cos2x\;dx\]

Answer 10

After splitting it into a couple of integrals like that, I integrated the first one.

Answer 11

so the first would be (1/4)x^2, but wouldn't the 1/2 stay with the second integral?

Answer 12

Yes it would, I see I made a silly mistake there, heh.

Answer 13

\[\Large\bf\sf \frac{1}{4}x^2-\frac{1}{2}\int\limits\limits x \cos 2x\;dx\]

Answer 14

so would u=x, du=dx, dv=cos2x, and v=...well im stuck here

Answer 15

Ya looks good so far.

Answer 16

Use advanced guessing. Let's not bother with a substitution to solve for v.

Integrating cos2x should give you `something like` sin2x, right?

Answer 17

yup

Answer 18

If we take the derivative of sin2x, we end up with an extra factor of 2 because of the inner function.

So our answer should have a factor of 1/2 to compensate for the fact that we don't want a 2 there.

When you differentiate, you gain a factor of that coefficient on x.
You're probably familiar with that.

When you integrate, you lose a factor of ( or divide out) that coefficient on x.

Answer 19

Since the inner function is 2x, and we're integrating, we'll end up dividing by 2.

\[\Large\bf\sf dv=\int\limits \cos2x\;dx, \qquad\qquad\qquad v=\frac{1}{2}\sin2x\]Mmmm does that explanation make a little sense maybe? D:
It's a little weird with integration, hard to get used to.

Answer 20

(1/2)xsin2x -∫(1/2)sin2x dx right?

Answer 21

Ya looks good.

Answer 22

(1/2)xsin2x- ....

im drawing a blank on that integral

Answer 23

Your first piece from `by parts` isn't contained in an integral, right?

Answer 24

right

Answer 25

\[\Large\bf\sf \color{#DD4747 }{\frac{1}{4}x^2-\left(\frac{1}{2}x \sin2x\color{black}{-\frac{1}{2}\int\limits \sin2x\;dx}\right)}\]So the pink parts are `done`.
Are you having trouble with this last term now?

Answer 26

yeah im drawing a blank on how to do it

Answer 27

You can do a u-sub if you're REALLY stuck.
But I would recommend trying to get comfortable with integrals like this.

Example:\[\Large\bf\sf \int\limits e^{2x}\;dx \quad=\quad \frac{1}{2}e^{2x}\]Just divide by the coefficient on x, instead of multiplying.\[\Large\bf\sf \int\limits \sec^24x\;dx\quad=\quad \frac{1}{4}\tan 4x\]
Understanding how this pattern works? :o
Integrate normally, then divide by coefficient.

Integral of sinx?

Answer 28

cosx

Answer 29

That's the derivative of sinx

Answer 30

right lol -cosx

Answer 31

Ok so the integral of sin2x will be -cos2x then we divide by the coefficient on x also.


\[\Large\bf\sf \int\limits \sin2x\;dx\quad=\quad -\frac{1}{2}\cos2x\]

Answer 32

ohhh ok

Answer 33

ok i think i got it from here. thanks

Answer 34

cool c: