How do you integrate #int sin^2(x) dx# using integration by parts?

Question

How do you integrate #int  sin^2(x) dx# using integration by parts?

Scarlett Allison · Accepted Answer

#intsin^2(x)dx=x/2-sin(2x)/4+c#,
where #c# is the constant of integration.

#intsin^2(x)dx=intfrac{d}{dx}(x)sin^2(x)dx# #=xsin^2(x)-intxfrac{d}{dx}(sin^2(x))dx# #=xsin^2(x)-intx(2sin(x)cos(x))dx# #=xsin^2(x)-intxsin(2x)dx# #=xsin^2(x)+1/2intxfrac{d}{dx}(cos(2x))dx# #=xsin^2(x)+1/2[xcos(2x)-intfrac{d}{dx}(x)cos(2x)dx]# #=xsin^2(x)+1/2xcos(2x)-1/2intcos(2x)dx# #=xsin^2(x)+1/2xcos(2x)-sin(2x)/4+c#, where #c# is the constant of integration. #=x/2(cos(2x)+2sin^2(x))-sin(2x)/4+c# #=x/2-sin(2x)/4+c#

Isabella Ackerman · Answer

undefined To integrate $ \int \sin^2(x) \, dx $ using integration by parts, we use the formula:

$$ \int u \, dv = uv - \int v \, du $$

Let's choose:

$$ u = \sin(x) $$
$$ dv = \sin(x) \, dx $$

Then, we find:

$$ du = \cos(x) \, dx $$
$$ v = -\cos(x) $$

Now, we apply the integration by parts formula:

$$ \int \sin^2(x) \, dx = -\sin(x) \cos(x) - \int (-\cos(x))(\cos(x) \, dx) $$

$$ = -\sin(x) \cos(x) + \int \cos^2(x) \, dx $$

We can rewrite $ \cos^2(x) $ as $ 1 - \sin^2(x) $:

$$ \int \cos^2(x) \, dx = \int (1 - \sin^2(x)) \, dx $$

$$ = \int dx - \int \sin^2(x) \, dx $$

$$ = x - \int \sin^2(x) \, dx $$

Now, let's solve for $ \int \sin^2(x) \, dx $:

$$ 2\int \sin^2(x) \, dx = -\sin(x) \cos(x) + x $$

$$ \int \sin^2(x) \, dx = -\frac{1}{2}\sin(x) \cos(x) + \frac{1}{2}x + C $$

Where $ C $ is the constant of integration.