Moved - integral problem please help me :(
- Thread starter yeliz
- Start date
how do you integrate:
(1-t^3)^25.t^5 dt
thanks
Is it:
\(\displaystyle \int(1-t^3)^{25t^5}dt\)
or
\(\displaystyle \int(1-t^3)^{25}t^5dt\)
or
something else??
the second one
thank you !
Hint: Let u = 1-t^3 and recognize that t^5 is the same at (t^2)(t^3).
Hello, yeliz!
By parts: .\(\displaystyle \begin{Bmatrix}u &=& t^3 && dv &=& (1-t^3)^{25}(t^2\,dt) \\ du &=& 3t^2\,dt && v &=& \text{-}\frac{1}{78}(1-t^3)^{26} \end{Bmatrix}\)
\(\displaystyle \displaystyle I \;=\;\text{-}\tfrac{1}{78}t^3(1-t^3)^{26} - \int \left(\text{-}\tfrac{1}{78}(1-t^3)^{26}\right)\left(3t^2\,dt\right) \)
\(\displaystyle \displaystyle I \;=\;\text{-}\tfrac{1}{78}t^3(1-t^3)^{26} + \tfrac{1}{78}\int(1-t^3)^{26}(3t^2\,dt) \)
\(\displaystyle \displaystyle I \;=\;\text{-}\tfrac{1}{78}t^3(1-t^3)^{26} - \tfrac{1}{78\cdot27}(1-t^3)^{27} + C\)
\(\displaystyle I \;=\;\text{-}\frac{1}{78\cdot27}(1-t^3)^{26}\big[27t^3+ (1-t^3)\big] + C\)
\(\displaystyle I \;=\;\text{-}\frac{1}{2106}(1-t^3)^{26}(26t^3 + 1) + C\)
By parts: .\(\displaystyle \begin{Bmatrix}u &=& t^3 && dv &=& (1-t^3)^{25}(t^2\,dt) \\ du &=& 3t^2\,dt && v &=& \text{-}\frac{1}{78}(1-t^3)^{26} \end{Bmatrix}\)
\(\displaystyle \displaystyle I \;=\;\text{-}\tfrac{1}{78}t^3(1-t^3)^{26} - \int \left(\text{-}\tfrac{1}{78}(1-t^3)^{26}\right)\left(3t^2\,dt\right) \)
\(\displaystyle \displaystyle I \;=\;\text{-}\tfrac{1}{78}t^3(1-t^3)^{26} + \tfrac{1}{78}\int(1-t^3)^{26}(3t^2\,dt) \)
\(\displaystyle \displaystyle I \;=\;\text{-}\tfrac{1}{78}t^3(1-t^3)^{26} - \tfrac{1}{78\cdot27}(1-t^3)^{27} + C\)
\(\displaystyle I \;=\;\text{-}\frac{1}{78\cdot27}(1-t^3)^{26}\big[27t^3+ (1-t^3)\big] + C\)
\(\displaystyle I \;=\;\text{-}\frac{1}{2106}(1-t^3)^{26}(26t^3 + 1) + C\)
Or another way:
\(\displaystyle \int(1-t^3)^{25}t^5dt=\int(1-t^3)^{25}t^3t^2dt\)
Let \(\displaystyle u=1-t^3\)
Then \(\displaystyle du=-3t^2dt\)
Thus \(\displaystyle t^2dt=-\frac{du}{3}\)
And \(\displaystyle t^3=1-u\)
Therefore:
\(\displaystyle \int(1-t^3)^{25}t^3t^2dt=\int(u)^{25}(1-u)(-\frac{du}{3})\)
\(\displaystyle =-\frac{1}{3}\int(u^{25}-u^{26})du\)
\(\displaystyle =-\frac{1}{3}(\frac{u^{26}}{26}-\frac{u^{27}}{27})+C\)
\(\displaystyle =\frac{1}{3}(\frac{u^{27}}{27}-\frac{u^{26}}{26})+C\)
\(\displaystyle =\frac{1}{3}[\frac{(1-t^3)^{27}}{27}-\frac{(1-t^3)^{26}}{26}]+C\)
Oh and be afraid, be very afraid, of Wolfram's solution. I just took a look and you got to be kidding me.
\(\displaystyle \int(1-t^3)^{25}t^5dt=\int(1-t^3)^{25}t^3t^2dt\)
Let \(\displaystyle u=1-t^3\)
Then \(\displaystyle du=-3t^2dt\)
Thus \(\displaystyle t^2dt=-\frac{du}{3}\)
And \(\displaystyle t^3=1-u\)
Therefore:
\(\displaystyle \int(1-t^3)^{25}t^3t^2dt=\int(u)^{25}(1-u)(-\frac{du}{3})\)
\(\displaystyle =-\frac{1}{3}\int(u^{25}-u^{26})du\)
\(\displaystyle =-\frac{1}{3}(\frac{u^{26}}{26}-\frac{u^{27}}{27})+C\)
\(\displaystyle =\frac{1}{3}(\frac{u^{27}}{27}-\frac{u^{26}}{26})+C\)
\(\displaystyle =\frac{1}{3}[\frac{(1-t^3)^{27}}{27}-\frac{(1-t^3)^{26}}{26}]+C\)
Oh and be afraid, be very afraid, of Wolfram's solution. I just took a look and you got to be kidding me.
Last edited:
D
Deleted member 4993
Guest
Wolfram solution:
integral (1-t^3)^25 t^5 dt
= -t^81/81+(25 t^78)/78-4 t^75+(575 t^72)/18-(550 t^69)/3+805 t^66-(25300 t^63)/9+(24035 t^60)/3-18975 t^57+(2042975 t^54)/54 -
(192280 t^51)/3 + (185725 t^48)/2-(1040060 t^45)/9+(371450 t^42)/3-(1485800 t^39)/13+(817190 t^36)/9-(185725 t^33)/3+(72105 t^30)/2 -
(480700 t^27)/27+ (44275 t^24)/6-2530 t^21+(6325 t^18)/9-(460 t^15)/3+25 t^12-(25 t^9)/9+t^6/6+constant
Well it is correct but clumsy - what would you expect from a free-software!!
integral (1-t^3)^25 t^5 dt
= -t^81/81+(25 t^78)/78-4 t^75+(575 t^72)/18-(550 t^69)/3+805 t^66-(25300 t^63)/9+(24035 t^60)/3-18975 t^57+(2042975 t^54)/54 -
(192280 t^51)/3 + (185725 t^48)/2-(1040060 t^45)/9+(371450 t^42)/3-(1485800 t^39)/13+(817190 t^36)/9-(185725 t^33)/3+(72105 t^30)/2 -
(480700 t^27)/27+ (44275 t^24)/6-2530 t^21+(6325 t^18)/9-(460 t^15)/3+25 t^12-(25 t^9)/9+t^6/6+constant
Well it is correct but clumsy - what would you expect from a free-software!!
HallsofIvy
Elite Member
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- Jan 27, 2012
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How is this a "pre-algebra problem"?