# Thread: Integral of (tanx)^3 * (secx)^88

1. ## Integral of (tanx)^3 * (secx)^88

$\displaystyle \int tan^3xsec^{88}x dx$

$\displaystyle =\int tan^2xsec^{88}x tanx dx$

$\displaystyle =\int (sex^2x-1)(sec^{88}x) tanx dx$

$\displaystyle u=sec^2x, du=tanx$

$\displaystyle =\int (u-1)u^{44} du$

$\displaystyle =\int u^45-u^44 du$

$\displaystyle =\frac{1}{46}u^{46}-\frac{1}{45}u^{45}$

$\displaystyle =\frac{1}{46}(sec^2x)^{46}-\frac{1}{45}(sec^2x)^{45}$

$\displaystyle =\frac{1}{46}sec^{94}x-\frac{1}{45}sec^{90}x$

Is this done correctly?

2. Originally Posted by circuscircus
$\displaystyle u=sec^2x, du=tanx$
You've got it backward:
$\displaystyle \frac{d}{dx}tan(x) = sec^2(x)$
not the other way around.

And then there's that comment about the sex...

-Dan

3. $\displaystyle \int tan^3xsec^{88}x dx$

$\displaystyle =\int tan^2xsec^{88}x tanx dx$

$\displaystyle =\int (sec^2x-1)(sec^{87}x) secxtanx dx$

$\displaystyle u=secx, du=secxtanx$

$\displaystyle =\int (u^2-1)u^87 du$

$\displaystyle =\int u^89-u^87 du$

$\displaystyle =\frac{1}{90}u^90-\frac{1}{88}u^88$

$\displaystyle =\frac{1}{90}sec^{90}x-\frac{1}{88}sec^{88}x$

So like this?

4. Originally Posted by circuscircus
$\displaystyle \int tan^3xsec^{88}x dx$

$\displaystyle =\int tan^2xsec^{88}x tanx dx$

$\displaystyle =\int (sec^2x-1)(sec^{87}x) secxtanx dx$

$\displaystyle u=secx, du=secxtanx$

$\displaystyle =\int (u^2-1)u^87 du$

$\displaystyle =\int u^89-u^87 du$

$\displaystyle =\frac{1}{90}u^90-\frac{1}{88}u^88$

$\displaystyle =\frac{1}{90}sec^{90}x-\frac{1}{88}sec^{88}x$

So like this?
It looks right to me, except for some LaTeX coding errors.

Oh yeah, and that pesky arbitrary constant...

-Dan