# Thread: Im struggling with integrals

1. ## Im struggling with integrals

Integral 2x^2 cos (1/2 x^3) dx

I dont understand!! can any body help?

2. Originally Posted by mdo112
Integral 2x^2 cos (1/2 x^3) dx

I dont understand!! can any body help?
$\displaystyle \int{2x^2\cos\left(\frac{1}{2}x^3\right)dx}$

We will use a method called u-substitution

If we let $\displaystyle u=x^3$

we would then have that

$\displaystyle du=3x^2dx\Rightarrow\frac{du}{3}=x^2dx$
So now lets replace each

$\displaystyle \int{2x^2\cos\left(\frac{1}{2}x^3\right)}dx\iff{2\ int\cos\left(\frac{1}{2}u\right)}\frac{du}{3}=\fra c{2}{3}\int\cos\left(\frac{1}{2}u\right)du$

3. Originally Posted by mdo112
Integral 2x^2 cos (1/2 x^3) dx

I dont understand!! can any body help?
Let $\displaystyle u=\frac{1}{2}x^3 \implies du =\frac{3}{2}x^2dx \iff \frac{2}{3}du=x^2dx$

So we get

$\displaystyle 2\int x^2\cos\left( \frac{1}{2}x^3\right)dx =2 \int \cos(u)\left( \frac{2}{3}du\right) =\frac{4}{3}\int \cos(u)du$

Now we can take the antiderivative to get

$\displaystyle \frac{4}{3}\sin(u )+C$ We can now sub back in what u is in terms of x to get

$\displaystyle \frac{4}{3}\sin\left( \frac{1}{2}x^3\right)+C$

4. Originally Posted by TheEmptySet
Let $\displaystyle u=\frac{1}{2}x^3 \implies du =\frac{3}{2}x^2dx \iff \frac{2}{3}du=x^2dx$

So we get

$\displaystyle 2\int x^2\cos\left( \frac{1}{2}x^3\right)dx =2 \int \cos(u)\left( \frac{2}{3}du\right) =\frac{4}{3}\int \cos(u)du$

Now we can take the antiderivative to get

$\displaystyle \frac{4}{3}\sin(u )+C$ We can now sub back in what u is in terms of x to get

$\displaystyle \frac{4}{3}\sin\left( \frac{1}{2}x^3\right)+C$
I really love seeing integrals done in two different ways..its neat everyone has their own style

5. Let $\displaystyle u=\frac{1}{2}x^3\to du=\frac{3}{2}x^2\;dx$

$\displaystyle \frac{4}{3} \int \cos u \;du= \frac{4}{3} \sin u+C$

Therefore, $\displaystyle 2x^2\cos\left(\frac{1}{2}x^3\right)=\frac{4}{3}\si n\left(\frac{1}{2}x^3\right)+C$

Darn...to slow