U-Substitution?

[Vinny]

How do I evaluate the integral ((x^2)sinx)/(5+x^6) dx on the period of -pie/4 to pie/4? I tried breaking them apart and integrating the numerator and denominator but i didn't get very far.

I also had trouble with two more problems I don't think are as hard, but i felt like I was missing a golden link. If f is continuos and on the interval of [0,54]f(x)dx=84, find f(6x)dx on [0,9].

The last problem I could not figure out was the indefinate integral (3+x)/(36+x^2)dx. I have five possible solutions for all my problems but I really couldn't get an answer.
Thank you for taking time to answer these questions.

 

[soroban]

Hello, Vinny!

The last one quite simple . . . Make two fractions.

\(\displaystyle \L\int\frac{x\,+\,3}{x^2\,+\,36}\,dx\;=\;\int\left(\frac{x}{x^2\,+\,36}\,+\,\frac{3}{x^2\,+\,3}\right)\,dx\)

The first integral is of the \(\displaystyle ln\) form . . . let \(\displaystyle u\,=\,x^2\,+\,3\)

The second is of the \(\displaystyle arctan\) form.

If \(\displaystyle f\) is continuous on the interval of \(\displaystyle [0,\,54]\)

and \(\displaystyle \L\int^{\;\;\;54}_0 f(x)\,dx\:=\:84\)

find \(\displaystyle \L \int^{\;\;\;9}_0 f(6x)\,dx\)

Suppose \(\displaystyle f(x)\) looks like this:

        |     *   *
        | *         *
        *             *           54
    ----+---------------*---------+-
        |                   *     :
        |                         *

Then \(\displaystyle f(6x)\) looks like this:

        |  * *
        |*    *
        *      *     9
    ----+-------*----+---
        |         *  :
        |            *

There has been a horizontal compression by a factor of 6.
The ordinates (y-values) are still the same.
Hence, the area is one-sixth of the area under \(\displaystyle f(x)\).

My answer is: \(\displaystyle \,\frac{1}{6}\,\times\,84\:=\;14\)