Re: Evaluate the integral

Re: Evaluate the integral

Quote:

Originally Posted by

**MINOANMAN** Choose A. : π/48

I know what the answer is -__-"

I wanted help on approaching how to solve something like this....

As on any math forum, giving THE answer to a question never really helps anyone...

Re: Evaluate the integral

Re: Evaluate the integral

Do each part separately: $\displaystyle \int_{-2}^2 \frac{x^5dx}{x^6+ 64}+ \int_{-2}^2\frac{x^2 dx}{x^6+ 64}+\int_{-2}^2 \frac{4xdx}{x^6+ 64}+ \int_{-2}^2\fra{sin(x) dx}{x^6+ 64}$

In the first one, let $\displaystyle u= x^6+ 64$. In the second, let $\displaystyle u= x^3$. In the third, let $\displaystyle u= x^2$. The last is trivial because the integrand is an odd function.

Re: Evaluate the integral

Quote:

Originally Posted by

**HallsofIvy** Do each part separately: $\displaystyle \int_{-2}^2 \frac{x^5dx}{x^6+ 64}+ \int_{-2}^2\frac{x^2 dx}{x^6+ 64}+\int_{-2}^2 \frac{4xdx}{x^6+ 64}+ \int_{-2}^2\fra{sin(x) dx}{x^6+ 64}$

In the first one, let $\displaystyle u= x^6+ 64$. In the second, let $\displaystyle u= x^3$. In the third, let $\displaystyle u= x^2$. The last is trivial because the integrand is an odd function.

Thanks! Just what I was looking for!

Re: Evaluate the integral

Quote:

Originally Posted by

**HallsofIvy** Do each part separately: $\displaystyle \int_{-2}^2 \frac{x^5dx}{x^6+ 64}+ \int_{-2}^2\frac{x^2 dx}{x^6+ 64}+\int_{-2}^2 \frac{4xdx}{x^6+ 64}+ \int_{-2}^2\fra{sin(x) dx}{x^6+ 64}$

In the first one, let $\displaystyle u= x^6+ 64$. In the second, let $\displaystyle u= x^3$. In the third, let $\displaystyle u= x^2$. The last is trivial because the integrand is an odd function.

The first and third are also odd functions.

- Hollywood