integration.. by reduction maybe?

[hgaon001]

integral of 12(tan^6(3x))dx

 

[soroban]

Hello, hgaon001!

Yes, we can reduce the integrand . . .

$\displaystyle 12\int\tan^6(3x)\,dx$

$\displaystyle \tan^6(3x) \:=\:\tan^4(3x)\tan^2(3x)$

. . . . . . $\displaystyle =\;\tan^4(3x)\overbrace{\left[\sec^2(3x) - 1\right]}$

. . . . . . $\displaystyle =\;\tan^4(3x)\sec^2(3x) - \tan^4(3x)$

. . . . . . $\displaystyle =\;\tan^4(3x)\sec^2(3x) - \overbrace{\tan^2(3x)\tan^2(3x)}$

. . . . . . $\displaystyle =\;\tan^4(3x)\sec^2(3x) - \tan^2(3x)\overbrace{\left[\sec^2(3x) - 1\right]}$

. . . . . . $\displaystyle =\;\tan^4(3x)\sec^2(3x) - \tan^2(3x)\sec^2(3x) + \tan^2(3x)$

. . . . . . $\displaystyle =\;\tan^4(3x)\sec^2(3x) - \tan^2(3x)\sec^2(3x) + \overbrace{\sec^2(3x) - 1}$


\(\displaystyle \text{We have: }\:12\bigg[\int\!\tan^4(3x)\sec^2(3x)\,dx \;- \int\!\tan^2(3x)\sec^2(3x)\,dx \;+ \int\!\sec^2(3x)\.dx \;- \int\!dx\bigg]\)

Can you finish it now?

 

[BigGlenntheHeavy]

Another way: If you look in the back of your Calc. book, it sould have a list of commonly used integrals.

$\displaystyle The \ one \ we \ are \ interested \ in \ is; \ \int tan^{n}(u)du \ = \ \frac{tan^{n-1}(u)}{n-1} \ - \ \int tan^{n-2}(u)du.$

$\displaystyle Hence, \ 12\int tan^{6}(3x)dx \ = \ 4\int tan^{6}(u)du, \ letting \ u \ = \ 3x, \ we \ get\frac{du}{3} \ = \ dx.$

$\displaystyle Ergo, \ 4\int tan^{6}(u)du \ = \ 4[\frac{tan^{5}(u)}{5} \ - \ \int tan^{4}(u)du]$

$\displaystyle Now, \ \int tan^{4}(u)du \ = \ \frac{tan^{3}(u)}{3} \ - \ \int tan^{2}(u)du \ and$

$\displaystyle \int tan^{2}(u)du \ = \ tan(u)-\int du \ = \ tan(u)-u+C.$

$\displaystyle Putting \ it \ all \ together \ we \ get: \ 4\int tan^{6}(u)du \ = \ 4[\frac{tan^{5}(u)}{5}-\frac{tan^{3}(u)}{3}+tan(u)-u]+C$

$\displaystyle Ergo: \ 12\int tan^{6}(3x)dx \ = \ \frac{4tan^{5}(3x)}{5}-\frac{4tan^{3}(3x)}{3}+4tan(3x)-12x+C, \ u \ = \ 3x, \ du \ = \ 3dx.$


$\displaystyle Check: \ D_x[\frac{4tan^{5}(3x)}{5}-\frac{4tan^{3}(3x)}{3}+4tan(3x)-12x+C] \ = \ 12tan^{6}(3x).$

$\displaystyle The \ above \ check \ would \ be \ a \ good \ exercise \ in \ your \ derivative \ acumen.$