# Math Help - Trigonometric integration

1. ## Trigonometric integration

Here's the question :

Evaluate the foll.intgral

[Int] 2 x^2 . cos (1/2 x^3 ) dx

I tried to do it by u-substitution and got this answer :

[4/3 x^3 . 4 sin (1/2 x^3 ) . 2 x^-2] + c

This does not look right.
Can someone pls show me how to do it?

2. Hi

Originally Posted by vdemas
Here's the question :

Evaluate the foll.intgral

[Int] 2 x^2 . cos (1/2 x^3 ) dx

I tried to do it by u-substitution and got this answer :

[4/3 x^3 . 4 sin (1/2 x^3 ) . 2 x^-2] + c

This does not look right.
Can someone pls show me how to do it?
$I=\int 2x^2 \cos \left(\frac 12 \cdot x^3 \right) dx$

That's correct, make a substitution : $u=\frac 12 x^3$

--> $du=\frac 32 x^2 dx \implies dx=\frac 2{3x^2} du$

$\implies I=\int 2{\color{red}x^2} \cdot \cos u \cdot \frac 2{3{\color{red}x^2}} du=\frac 43 \int \cos u du$

3. well if you dont want to use substitution you can recognize from the differentiation of sine functions:

$
\frac {d}{d\theta}sin(\theta) = cos(\theta)
$

$
\frac {d}{d\theta}sin(\theta ^2) = 2\theta \cdot cos(\theta ^2)
$

now lets look at your case:
$
I=\int 2x^2 \cos \left(\frac 12 \cdot x^3 \right) dx
$

looking at the trigonometry what do you differentiate to obtain:
$
\cos \left(\frac 12 \cdot x^3 \right)
$

well you can see that $\sin \left(\frac 12 \cdot x^3 \right)$ will give you the required cos function:

$
\frac{d}{dx} \sin \left(\frac 12 \cdot x^3 \right) = \frac 32 x^2\cos \left(\frac 12 \cdot x^3 \right)
$

From here we can see that differentiating $\sin \left(\frac 12 \cdot x^3 \right)$ will give us almost the required answer.
Multiplying both side with 4/3 gives:

$
\frac 43 \cdot \frac{d}{dx} \sin \left(\frac 12 \cdot x^3 \right) = \frac43 \cdot \frac 32 x^2\cos \left(\frac 12 \cdot x^3 \right)
$

$
\frac{d}{dx} \frac 43 \sin \left(\frac 12 \cdot x^3 \right) = 2 x^2\cos \left(\frac 12 \cdot x^3 \right)
$

Now we just integrate both sides to obtain the required question:
$
\int \frac{d}{dx} \frac 43 \sin \left(\frac 12 \cdot x^3 \right) = \int 2 x^2\cos \left(\frac 12 \cdot x^3 \right)
$

$
\int 2 x^2\cos \left(\frac 12 \cdot x^3 \right)= \frac 43 \sin \left(\frac 12 \cdot x^3 \right)
$