integration by parts help please - sin^2x?

so use u = sin x and dv = sin x dx...and remember that cos² x = 1 - sin² x

Hi. Your question stomped me for a while, so when I signed out last night, I had to get a pen and paper and work it out. I know how to do it now. I was thinking on the same lines as Ted before, but I'm sure you know what happens when you do what he suggested: you just get an identity.

Anyway, the secret lies with the identity

sin^2 (x) = 1 - cos^2 (x)

and the factoring

1 - cos^2 (x) = (1 + cos (x)) (1 - cos (x))

Now, going back, we have

int {sin^2 (x)} dx = int {1 - cos^2 (x)} dx

int {sin^2 (x)} dx = int {(1 + cos (x)) (1 - cos (x))} dx

Now, for integration by parts. We have

u = 1 + cos (x)

du = -sin (x)

dv = (1 - cos (x)) dx

v = x - sin (x)

Substituting according to the form of integration by parts, we have

int {sin^2 (x)} dx = (1 + cos (x)) (x - sin (x)) - int {(x - sin (x)) (-sin (x))} dx

int {sin^2 (x)} dx = (1 + cos (x)) (x - sin (x)) + int {x sin (x) - sin^2 (x)} dx

int {sin^2 (x)} dx = (1 + cos (x)) (x - sin (x)) + int {x sin (x)} dx - int {sin^2 (x)} dx

2 * int {sin^2 (x)} dx = (1 + cos (x)) (x - sin (x)) + int {x sin (x)} dx

Now, focusing on the remaining integral on the right side, we have

u = x

du = dx

dv = sin (x) dx

v = -cos (x)

int {x sin(x)} dx = x (-cos (x)) - int {(-cos (x)} dx

int {x sin(x)} dx = -x cos (x) + sin (x)

Substituting this into the original, we have

2 * int {sin^2 (x)} dx = (1 + cos (x)) (x - sin (x)) + -x cos (x) + sin (x)

2 * int {sin^2 (x)} dx = x - sin (x) + x cos (x) - cos (x) sin (x) - x cos (x) + sin (x) + C

Combining like terms, we have

2 * int {sin^2 (x)} dx = x - cos (x) sin (x) + C

Now, using an identity for sin (2x), which is

sin (2x) = 2 sin (x) cos (x)

sin (x) cos (x) = 1/2 sin (2x)

we now have

[2 * int {sin^2 (x)} dx = x - 1/2 sin (2x) + C] / 2

int {sin^2 (x)} dx = 1/2 x - 1/4 sin (2x) + C

Now, that is the integral of sin^2 (x), is it not?

Hope this helps!!!!!

∫ sin^2 x dx

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This is a tricky one but I'll try to explain it as best as I can. Have you learned tabular integration? It's much easier than using parts for this one, but I'll explain this one by using parts.

In essence, integration by parts involves the basic formula, ∫ u dv = uv - ∫ v du

Integrating by parts is used when the derivative is too complex to anti-differentiate using basic u substitution, so we will split the equation into a u to differentiate and a dv to anti differentiate. Usually the us get simpler and the dvs don't get more complicated. So this one is a bit tricky to choose a u and v for, but we will choose sin^2 x as our u because finding the derivative for that is easy, while we will use dx as our dv because the anti-differentation of dx is just x, talk about easy! :D

So using that, we have this:

u = sin^2(x)

du = ?

v = x

dv = dx

So for the derivative of sin^2 x, that's just using the product rule and chain rule. the exponent goes in front of the sin x, so we have 2 sin x, but we need to use chain rule and multiply by the derivative of sin x, which is cos x, so we now have:

u = sin^2 x

du = 2 sin x cos x

v = x

dv = dx

So we now plug in what we have into the equation I initially stated, uv - ∫ v du:

x sin^2 x - ∫ 2x sin x cos x

We can pull out the constant to make this less complicated.. but we are going to have to use a trig identity for the next step because we can't anti differentiate this.

∫ x sin x cos x , sin x cos x = 1/2 sin (2x)

= ∫ x sin (2x)

...which means we have to do another integration by parts within it. I know it's going to be confusing, but just bear with it.

u = x

du = dx

v = -1/2 cos (2x) *

dv = sin (2x) dx

* Think of this as the reverse chain rule. because when you find the derivative of cos 2x, you will have to use chain rule to multiply the 2 to the sin x. However, there's no two in front of sin 2x, so the half fixes that.

uv - ∫ v du

-(1/2)x cos (2x) - ∫ (-1/2) cos (2x)

We pull out the constant in front of the integral, but note the negative 1/2 cancels with the negative 1 outside, so the integral becomes positive:

(1/2) ∫ cos 2x dx

We will use u substitution, u = 2x; du = 2dx (divide each side by 2, so we can then pull out another 1/2 of the integral)

(1/4) ∫ cos u du , which becomes (1/4) sin (2x), and we plug that back into the equation:

-(1/2)x cos (2x) - (1/4 sin (2x))

return to the where we left off when we started integrating by parts the 2nd time... if you don't remember we had this:

x sin^2 x - ∫ 2x sin x cos x

And we had to integrate by parts for that integral, so now that we have that answer, we can plug in what we have:

x sin^2 x - (-(1/2)x cos (2x) - ((1/4) sin (2x)))

Which becomes when we distribute the negative:

x sin^2 x + 1/2x cos (2x) + (1/4) sin (2x)

And the final step, which we have to always remember, we have to add the + C at the end of the equation.

x sin^2 x + 1/2x cos (2x) + (1/4) sin (2x) + C

I'm not sure if it's right.. but I think you can get the concept. I hope I at least helped somehow.