Question

what is the integral 1/2 to 1 (cos(x^-2))/x^3 dx
and the other one is the integral 0 to 9 xsqrt(x^2+a^2)dx (a>0)

Answer 1

To solve these integrals, you will need to utilize integration by parts. \[\int\limits_{a}^{b} udv = uv - \int\limits_{a}^{b} vdu\]

Answer 2

So it's
\[\int_{1/2}^1\frac{1}{x^3}\cos\left(\frac{1}{x^2}\right)~dx\]
and
\[\int_0^9x\sqrt{x^2+a^2}~dx\]
right?

Answer 3

I'd like to point out that integration by parts isn't necessary. Substitutions would work just fine. But if that's what the instructions say...

Answer 4

For the first one, integration by parts is necessary, since it's a polynomial multiplied by a trig function.

Answer 5

Substituting \(u=\dfrac{1}{x^2}\) works just fine. You have \(du=-\dfrac{2}{x^3}~dx\), or \(-\dfrac{1}{2}du=\dfrac{1}{x^3}~dx\), so the integral becomes
\[-\frac{1}{2}\int_4^1\cos u~du=\frac{1}{2}\int_1^4\cos u~du\]
I'm not saying IBP isn't a viable method, it's just that substitution is a generally simpler one.