Hello

In this post $\displaystyle integralof$ is the integration sign and $\displaystyle th$ is theta.

1) Integrate $\displaystyle x(2x^2-3)^5$

Let $\displaystyle u = x, du = 1$

$\displaystyle dv = (2x^2-3)^5, v = (2x^2-3)^6/24x$

$\displaystyle I = x(2x^2-3)^6/24x - integralof(2x^2-3)^6/24x dx$

I do not know how to carry this on. The answer is $\displaystyle (2x^2-3)^6/24 + k$

2) Substituting a trigonometric function for x, integrate $\displaystyle x^2/root(4-x^2)$

Let$\displaystyle x = 2sin(th)$

$\displaystyle I = integrate(4sin^2(th))/4cos^2(th) * 2cos(th) d(th))$

The terms cancel

$\displaystyle I = integralof(4sin^2(th) d(th))$

$\displaystyle I = integralof(2 - 2cos(2th) d(th))$

$\displaystyle I = 2th - sin(2th) + C$

$\displaystyle th = sin^-1(x/2)$

$\displaystyle :. I = 2sin^-1(x/2) - sin(2th)$

How do I manipulate $\displaystyle sin(2th)$ in terms of $\displaystyle x$?

The answer is $\displaystyle 2sin^-1(x/2) - (x/2)root(4-x^2) + k$

Thank you