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?
$\displaystyle I=\int 2x^2 \cos \left(\frac 12 \cdot x^3 \right) dx$

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

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

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

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

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

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

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

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

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

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

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

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

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