How do you integrate #int x^2 cos x^2 dx # using integration by parts?

Question

How do you integrate #int x^2   cos  x^2    dx  # using integration by parts?

Emma Aldridge · Accepted Answer

#int x^2cos^2xdx = x^3/6 +(x^2sin(2x))/4 + (xcos(2x))/4 -1/8sin2x +C#

When we integrate by parts a function of the form:

#x^nf(x)#

we normally choose #x^n# as the integral part and #f(x)# as the differential part, so that in the resulting integral we have #x^(n-1)#

In this case however

#cos^2xdx#

is not the differential of an «easy» function, so we first reduce the degree of the trigonometric function using the identity:

#cos^2x = (1+cos(2x))/2#

#int x^2cos^2xdx = int x^2(1+cos(2x))/2dx = int x^2/2dx +int x^2/2 cos(2x)dx#

Now we can solve the first integral directly:

#int x^2/2dx = x^3/6#

and the second by parts:

#int x^2/2 cos(2x)dx = 1/4 int x^2 d(sin2x) = (x^2sin(2x))/4 -1/2 int xsin(2x)dx#

and again:

#1/2 int xsin(2x)dx = -1/4 int xd(cos2x) = -(xcos(2x))/4 + 1/4 int cos2xdx = -(xcos(2x))/4 +1/8sin2x +C#

Putting it together:

#int x^2cos^2xdx = x^3/6 +(x^2sin(2x))/4 + (xcos(2x))/4 -1/8sin2x +C#

Emilia Aguilar · Answer

undefined To integrate $\int x^2 \cos(x^2) \, dx$ using integration by parts, we'll choose $u = x^2$ and $dv = \cos(x^2) \, dx$. Then, we have:

$$du = 2x \, dx$$
$$v = \int \cos(x^2) \, dx$$

To find $v$, we need to make a substitution. Let $w = x^2$, then $dw = 2x \, dx$. This gives us:

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

Now we can apply the integration by parts formula:

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

Substituting our values:

$$
\int x^2 \cos(x^2) \, dx = \frac{1}{2}x^2 \sin(x^2) - \int \left(\frac{1}{2} \sin(x^2)\right) \cdot (2x \, dx)
$$

Simplifying the integral:

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

Now we can use another substitution for the remaining integral. Let $z = x^2$, then $dz = 2x \, dx$, giving:

$$
= \frac{1}{2}x^2 \sin(x^2) - \frac{1}{4} \int \sin(z) \, dz
$$

Integrating $\int \sin(z) \, dz$ gives $-\frac{1}{4} \cos(z)$. Substituting back:

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

So, $\int x^2 \cos(x^2) \, dx = \frac{1}{2}x^2 \sin(x^2) - \frac{1}{4} \cos(x^2) + C$.