Thread: Integration by Substition

1. Integration by Substition

I find myself sitting at my desk, pondering why I decided to take Calc 2. Anyway, I'm having trouble with simple integration by substitution, which does not bode well for the rest of my semester.

Find the Following Integral: t^3(t^4-2)^5dt
So far I have done
u=t^4-2
du=4t^3dt

So i have ∫(u)^5(du/4)dt=1/4 ∫(t^4-2)^5*4t^3dt
simplifying...
1/4((t^4-2)^6)/6*t^4
Giving me a final (wrong) answer of
((t^4-2)^6*t^4)/24
What am I doing wrong here? Any help is extremely appreciated

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2. I hope that it helps. Sorry for the size being small.

3. Originally Posted by john84
I find myself sitting at my desk, pondering why I decided to take Calc 2. Anyway, I'm having trouble with simple integration by substitution, which does not bode well for the rest of my semester.

Find the Following Integral: t^3(t^4-2)^5dt
So far I have done
u=t^4-2
du=4t^3dt

So i have ∫(u)^5(du/4)dt=1/4 ∫(t^4-2)^5*4t^3dt
simplifying...
1/4((t^4-2)^6)/6*t^4
Giving me a final (wrong) answer of
((t^4-2)^6*t^4)/24
What am I doing wrong here? Any help is extremely appreciated

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$\int{t^3(t^4 - 2)^5\,dt} = \frac{1}{4}\int{4t^3(t^4 - 2)^5\,dt}$

Let $u = t^4 - 2$ so that $\frac{du}{dt} = 4t^3$.

The integral becomes

$\frac{1}{4}\int{u^5\,\frac{du}{dt}\,dt}$

$= \frac{1}{4}\int{u^5\,du}$

$= \frac{1}{4}\cdot\frac{1}{6}u^6 + C$

$= \frac{1}{24}u^6 + C$

$= \frac{1}{24}(t^4 - 2)^6 + C$.