Question

Use the substitution formula to evaluate the integral
(essentially finding the anti derivative)
t^5(1+t^6)^5 dt

I know this much
u= 1+t^6
du or g\prime = 6t^5
(doing online homework and i just need to really understand this step)
so when you evaluate t^5 dt you get 1/6 du

if you could show me each step to get there i would really appreciate it... this stuff isnt easy learning on your own with a crappy program

Answer 1

Let u= 1+t^6,
du = 6t^5 dt

∫t^5(1+t^6)^5 dt
= 1/6 (∫ 6t^5(1+t^6)^5 dt )
= 1/6 [∫ (1+t^6)^5 (6t^5dt) ]
= 1/6 ∫ u^5 du
= ...

Answer 2

so t^5 gets canceled out when you remove g\prime? and the 1/6 is simply removing the 6? how does that make it into a fraction?

Answer 3

Sorry, but I don't understand your questions...

Answer 4

show me exactly what you did between these two steps
∫t^5(1+t^6)^5 dt
= 1/6 (∫ 6t^5(1+t^6)^5 dt )

Answer 5

\[\int t^5(1+t^6)^5 dt\]\[=\frac{6}{6}\int t^5(1+t^6)^5 dt\]\[=\frac{1}{6}\int 6t^5(1+t^6)^5 dt\]
Got it?

Answer 6

Reason for it:
6t^5dt = du, when u = 1+t^6

Answer 7

yeah basically the whole point of that is to get the equation to equal the chain method? you take the derivative of u to multiply it back into the equation to get du. I understand that part

why is that fraction 1/6? in other words why does it end up on the other side of the anti-derivative sign in the first place the 6t^5 makes sense but the fraction seems to come from some rule that im missing

Answer 8

Because you can't just multiply the question by 6. If you do so, you change the question. Instead, we can multiply the integral by 1, which doesn't change the question.
So, I multiplied the integral by 6/6 = 1
Put 6 inside the integral. 1/6 as the remaining.

Answer 9

ahhhh so the sign acts like an = sign! thank you so much! sorry that was so difficult to get across! i just couldn't figure out how to ask that question!