1. ## Integral Help

I am stuck on this problem. I tried substituting u= cos^3(x^2), but it didnt work for me.
I am not asking for someone to do this for me I just need some help getting started.

2. ## Re: Integral Help

Originally Posted by nubshat

I am stuck on this problem. I tried substituting u= cos^3(x^2), but it didnt work for me.
I am not asking for osmeone to do this ofr me I just need soem help getting started.
$u = x^2$

$du = 2x \, dx$

$\frac{1}{2} \int_0^{\sqrt{\frac{\pi}{6}}} \cos(x^2)[\cos^2(x^2) - 1] \cdot 2x \, dx$

$-\frac{1}{2} \int_0^\frac{\pi}{6} \cos{u}(\sin^2{u}) \, du$

$t = \sin{u}$

$dt = \cos{u} \, du$

$-\frac{1}{2} \int_0^\frac{1}{2} t^2 \, dt$

3. ## Re: Integral Help

Hello, nubshat!

$\displaystyle \int^{\sqrt{\frac{\pi}{6}}}_0\left[\cos^3(x^2) - \cos(x^2)\right]x\,dx$

Note that: . $\cos^3(x^2) - \cos(x^2) \;=\;\cos(x^2)\left[\cos^2(x^2) - 1\right] \;=\;\cos(x^2)\left[-\sin^2(x^2)\right]$

The integral becomes: . $\displaystyle -\int\sin^2(x^2)\cdot \cos(x^2)\,x\,dx$

Let $u = \sin(x^2) \quad\Rightarrow\quad du = \cos(x^2)\cdot2x\cdot dx \quad\Rightarrow\quad \cos(x^2)\,x\,dx = \tfrac{1}{2}du$
. . When $x = 0\!:\;u = \sin(0^2) = 0.$
. . When $x = \sqrt{\tfrac{\pi}{6}}\!:\;u = \sin\tfrac{\pi}{6} = \tfrac{1}{2}$

Substitute: . $\displaystyle -\int^{\frac{1}{2}}_0 u^2\cdot\tfrac{1}{2}du \;=\;-\tfrac{1}{2}\int^{\frac{1}{2}}_0 u^2du \;=\;-\tfrac{1}{6}u^3\,\bigg]^{\frac{1}{2}}_0$

. . . . . . . . $=\;\left[-\tfrac{1}{6}\left(\tfrac{1}{2}\right)^3\right] - \left[-\tfrac{1}{6}(0^3)\right] \;=\;-\tfrac{1}{48}$