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integration.. by reduction maybe?
- Thread starter hgaon001
- Start date
Hello, hgaon001!
Yes, we can reduce the integrand . . .
\(\displaystyle \tan^6(3x) \:=\:\tan^4(3x)\tan^2(3x)\)
. . . . . . \(\displaystyle =\;\tan^4(3x)\overbrace{\left[\sec^2(3x) - 1\right]}\)
. . . . . . \(\displaystyle =\;\tan^4(3x)\sec^2(3x) - \tan^4(3x)\)
. . . . . . \(\displaystyle =\;\tan^4(3x)\sec^2(3x) - \overbrace{\tan^2(3x)\tan^2(3x)}\)
. . . . . . \(\displaystyle =\;\tan^4(3x)\sec^2(3x) - \tan^2(3x)\overbrace{\left[\sec^2(3x) - 1\right]}\)
. . . . . . \(\displaystyle =\;\tan^4(3x)\sec^2(3x) - \tan^2(3x)\sec^2(3x) + \tan^2(3x)\)
. . . . . . \(\displaystyle =\;\tan^4(3x)\sec^2(3x) - \tan^2(3x)\sec^2(3x) + \overbrace{\sec^2(3x) - 1}\)
\(\displaystyle \text{We have: }\:12\bigg[\int\!\tan^4(3x)\sec^2(3x)\,dx \;- \int\!\tan^2(3x)\sec^2(3x)\,dx \;+ \int\!\sec^2(3x)\.dx \;- \int\!dx\bigg]\)
Can you finish it now?
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
Senior Member
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- Mar 8, 2009
- Messages
- 1,577
Another way: If you look in the back of your Calc. book, it sould have a list of commonly used integrals.
\(\displaystyle The \ one \ we \ are \ interested \ in \ is; \ \int tan^{n}(u)du \ = \ \frac{tan^{n-1}(u)}{n-1} \ - \ \int tan^{n-2}(u)du.\)
\(\displaystyle Hence, \ 12\int tan^{6}(3x)dx \ = \ 4\int tan^{6}(u)du, \ letting \ u \ = \ 3x, \ we \ get\frac{du}{3} \ = \ dx.\)
\(\displaystyle Ergo, \ 4\int tan^{6}(u)du \ = \ 4[\frac{tan^{5}(u)}{5} \ - \ \int tan^{4}(u)du]\)
\(\displaystyle Now, \ \int tan^{4}(u)du \ = \ \frac{tan^{3}(u)}{3} \ - \ \int tan^{2}(u)du \ and\)
\(\displaystyle \int tan^{2}(u)du \ = \ tan(u)-\int du \ = \ tan(u)-u+C.\)
\(\displaystyle Putting \ it \ all \ together \ we \ get: \ 4\int tan^{6}(u)du \ = \ 4[\frac{tan^{5}(u)}{5}-\frac{tan^{3}(u)}{3}+tan(u)-u]+C\)
\(\displaystyle Ergo: \ 12\int tan^{6}(3x)dx \ = \ \frac{4tan^{5}(3x)}{5}-\frac{4tan^{3}(3x)}{3}+4tan(3x)-12x+C, \ u \ = \ 3x, \ du \ = \ 3dx.\)
\(\displaystyle Check: \ D_x[\frac{4tan^{5}(3x)}{5}-\frac{4tan^{3}(3x)}{3}+4tan(3x)-12x+C] \ = \ 12tan^{6}(3x).\)
\(\displaystyle The \ above \ check \ would \ be \ a \ good \ exercise \ in \ your \ derivative \ acumen.\)
\(\displaystyle The \ one \ we \ are \ interested \ in \ is; \ \int tan^{n}(u)du \ = \ \frac{tan^{n-1}(u)}{n-1} \ - \ \int tan^{n-2}(u)du.\)
\(\displaystyle Hence, \ 12\int tan^{6}(3x)dx \ = \ 4\int tan^{6}(u)du, \ letting \ u \ = \ 3x, \ we \ get\frac{du}{3} \ = \ dx.\)
\(\displaystyle Ergo, \ 4\int tan^{6}(u)du \ = \ 4[\frac{tan^{5}(u)}{5} \ - \ \int tan^{4}(u)du]\)
\(\displaystyle Now, \ \int tan^{4}(u)du \ = \ \frac{tan^{3}(u)}{3} \ - \ \int tan^{2}(u)du \ and\)
\(\displaystyle \int tan^{2}(u)du \ = \ tan(u)-\int du \ = \ tan(u)-u+C.\)
\(\displaystyle Putting \ it \ all \ together \ we \ get: \ 4\int tan^{6}(u)du \ = \ 4[\frac{tan^{5}(u)}{5}-\frac{tan^{3}(u)}{3}+tan(u)-u]+C\)
\(\displaystyle Ergo: \ 12\int tan^{6}(3x)dx \ = \ \frac{4tan^{5}(3x)}{5}-\frac{4tan^{3}(3x)}{3}+4tan(3x)-12x+C, \ u \ = \ 3x, \ du \ = \ 3dx.\)
\(\displaystyle Check: \ D_x[\frac{4tan^{5}(3x)}{5}-\frac{4tan^{3}(3x)}{3}+4tan(3x)-12x+C] \ = \ 12tan^{6}(3x).\)
\(\displaystyle The \ above \ check \ would \ be \ a \ good \ exercise \ in \ your \ derivative \ acumen.\)