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layd33foxx
02-26-2012, 12:19 PM
what did I do wrong:sad:

galactus
02-26-2012, 01:09 PM
I think the middle term should have a 5 instead of a 10.

10\left(\frac{1}{4}tan^{4}(x/10)-\frac{1}{2}tan^{2}(x/10)+ln(|sec(x/10)|)\right)

soroban
02-26-2012, 09:09 PM
Hello, layd33foxx!

Your answer should have had secants . . .


\displaystyle\text{Integrate: }\:I \;=\;\int\tan^5(\tfrac{x}{10})\,dx
Let u = \frac{x}{10} \quad\Rightarrow\quad du = \frac{1}{10}\:\!dx \quad\Rightarrow\quad dx = 10\:\!du

\displaystyle\text{Substitute: }\:I \;=\; 10\!\int\! \tan^5\!u\,du


Some gymnastics . . .

\tan^5\!u \;=\;(\tan^2\!u)^2\tan u \;=\;(\sec^2\!u - 1)^2\tan u \;=\;(\sec^4\!u - 2\sec^2\!u + 1)\tan u

. . . . . =\;\sec^4\!u\tan u - 2\sec^2\!u\tan u + \tan u

. . . . . =\;\sec^3\!u(\sec u\tan u) - 2\sec u(\sec u\tan u) + \tan u


\displaystyle\text{Then: }\:I \;=\;10 \bigg[\int \sec^3\!u(\sec u\tan u\,du) - 2\int\sec u(\sec u\tan u\,du) + \displaystyle\int\tan u\,du\bigg]


. . . . . .I \;=\;10\left[\frac{1}{4}\sec^4\!u - \sec^2\!u - \ln|\cos u|\right] + C

. . . . . .I \;=\;\frac{5}{2}\sec^4\!u - 10\sec^2\!u - 10\ln|\cos u| + C


Back-substitute:

. . I \;=\;\frac{5}{2}\sec^4(\frac{x}{10}) - 10\sec^2(\frac{x}{10}) - 10\ln|\cos(\frac{x}{10})| + C