what did I do wrong:sad:
L layd33foxx New member Joined Sep 24, 2011 Messages 26 Feb 26, 2012 #1 what did I do wrong:sad: Attachments 19.jpg 28.9 KB · Views: 2
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,216 Feb 26, 2012 #2 I think the middle term should have a 5 instead of a 10. \(\displaystyle 10\left(\frac{1}{4}tan^{4}(x/10)-\frac{1}{2}tan^{2}(x/10)+ln(|sec(x/10)|)\right)\)
I think the middle term should have a 5 instead of a 10. \(\displaystyle 10\left(\frac{1}{4}tan^{4}(x/10)-\frac{1}{2}tan^{2}(x/10)+ln(|sec(x/10)|)\right)\)
S soroban Elite Member Joined Jan 28, 2005 Messages 5,586 Feb 26, 2012 #3 Hello, layd33foxx! Your answer should have had secants . . . \(\displaystyle \displaystyle\text{Integrate: }\:I \;=\;\int\tan^5(\tfrac{x}{10})\,dx\) Click to expand... Let \(\displaystyle u = \frac{x}{10} \quad\Rightarrow\quad du = \frac{1}{10}\:\!dx \quad\Rightarrow\quad dx = 10\:\!du\) \(\displaystyle \displaystyle\text{Substitute: }\:I \;=\; 10\!\int\! \tan^5\!u\,du \) Some gymnastics . . . \(\displaystyle \tan^5\!u \;=\;(\tan^2\!u)^2\tan u \;=\;(\sec^2\!u - 1)^2\tan u \;=\;(\sec^4\!u - 2\sec^2\!u + 1)\tan u\) . . . . . \(\displaystyle =\;\sec^4\!u\tan u - 2\sec^2\!u\tan u + \tan u \) . . . . . \(\displaystyle =\;\sec^3\!u(\sec u\tan u) - 2\sec u(\sec u\tan u) + \tan u\) \(\displaystyle \displaystyle\text{Then: }\:I \;=\;10 \bigg[\int \sec^3\!u(\sec u\tan u\,du) - 2\int\sec u(\sec u\tan u\,du) +\) \(\displaystyle \displaystyle\int\tan u\,du\bigg] \) . . . . . .\(\displaystyle I \;=\;10\left[\frac{1}{4}\sec^4\!u - \sec^2\!u - \ln|\cos u|\right] + C \) . . . . . .\(\displaystyle I \;=\;\frac{5}{2}\sec^4\!u - 10\sec^2\!u - 10\ln|\cos u| + C \) Back-substitute: . . \(\displaystyle I \;=\;\frac{5}{2}\sec^4(\frac{x}{10}) - 10\sec^2(\frac{x}{10}) - 10\ln|\cos(\frac{x}{10})| + C\)
Hello, layd33foxx! Your answer should have had secants . . . \(\displaystyle \displaystyle\text{Integrate: }\:I \;=\;\int\tan^5(\tfrac{x}{10})\,dx\) Click to expand... Let \(\displaystyle u = \frac{x}{10} \quad\Rightarrow\quad du = \frac{1}{10}\:\!dx \quad\Rightarrow\quad dx = 10\:\!du\) \(\displaystyle \displaystyle\text{Substitute: }\:I \;=\; 10\!\int\! \tan^5\!u\,du \) Some gymnastics . . . \(\displaystyle \tan^5\!u \;=\;(\tan^2\!u)^2\tan u \;=\;(\sec^2\!u - 1)^2\tan u \;=\;(\sec^4\!u - 2\sec^2\!u + 1)\tan u\) . . . . . \(\displaystyle =\;\sec^4\!u\tan u - 2\sec^2\!u\tan u + \tan u \) . . . . . \(\displaystyle =\;\sec^3\!u(\sec u\tan u) - 2\sec u(\sec u\tan u) + \tan u\) \(\displaystyle \displaystyle\text{Then: }\:I \;=\;10 \bigg[\int \sec^3\!u(\sec u\tan u\,du) - 2\int\sec u(\sec u\tan u\,du) +\) \(\displaystyle \displaystyle\int\tan u\,du\bigg] \) . . . . . .\(\displaystyle I \;=\;10\left[\frac{1}{4}\sec^4\!u - \sec^2\!u - \ln|\cos u|\right] + C \) . . . . . .\(\displaystyle I \;=\;\frac{5}{2}\sec^4\!u - 10\sec^2\!u - 10\ln|\cos u| + C \) Back-substitute: . . \(\displaystyle I \;=\;\frac{5}{2}\sec^4(\frac{x}{10}) - 10\sec^2(\frac{x}{10}) - 10\ln|\cos(\frac{x}{10})| + C\)