How Do You Apply U-Substitution to ∫sin(x^5)dx?

[LearninDaMath]

Homework Statement

Use Part 2 of the Fundamental Theorem of Calculus to find the derivative.

\int_3^x sin(t^{5}) \, dt

Homework Equations

The Attempt at a Solution

I know the general idea of what I'm supposed to do as far as evaluate the indefinate integral and then do a subtraction of the upper limit and lower limits...but I can't even get to the point of finding the indefinate integral. (maybe it's that I just "think" I know what I'm supposed to do..)

\int_3^x sin(t^{5}) \, dt

I'm letting u = t^{5}

so du = 5t^{4}

then it looks like dt can be replaced by \frac{1}{5t^{4}}

so that \int_3^x \frac{1}{5t^4} sin(u) \, du

However, our professor has instructed that mixing variables within the integral is not allowed because it can't be evaluated.

So how to I do u-substitution on this integral?

 

[vela]

You don't. Note that the problem says to use the fundamental theorem of calculus. Don't ignore that bit of info.

 

[SammyS]

Homework Statement

Use Part 2 of the Fundamental Theorem of Calculus to find the derivative.
\int_3^x sin(t^{5}) \, dt

Homework Equations

The Attempt at a Solution

I know the general idea of what I'm supposed to do as far as evaluate the indefinate integral and then do a subtraction of the upper limit and lower limits...but I can't even get to the point of finding the indefinate integral. (maybe it's that I just "think" I know what I'm supposed to do..)
\int_3^x sin(t^{5}) \, dt
I'm letting u = t^{5}

so du = 5t^{4}

then it looks like dt can be replaced by \frac{1}{5t^{4}}

so that \int_3^x \frac{1}{5t^4} sin(u) \, du

However, our professor has instructed that mixing variables within the integral is not allowed because it can't be evaluated.

So how to I do u-substitution on this integral?

What vela said, plus ... you're to find the derivative.

 

[LearninDaMath]

oh, it says find the derivative...lol. I just saw the integral sign and began trying to integrate. I didn't even notice that there is a d/dt notation to the right of the integral clear as day. And so I figured the fundamental theorem being referred to was the property that is actually FTC1. Anyway, Thanks :) sin(x^5) duh. I turned a 30 second problem into a 20 minute headache. Thanks Vela and Sammy.

 

[SammyS]

oh, it says find the derivative...lol. I just saw the integral sign and began trying to integrate. I didn't even notice that there is a d/dt notation to the right of the integral clear as day. And so I figured the fundamental theorem being referred to was the property that is actually FTC1. Anyway, Thanks :) sin(x^5) duh. I turned a 30 second problem into a 20 minute headache. Thanks Vela and Sammy.

The limits od the integral are 3 and x, so the integral is a function of x, not t.

The derivative of this integral is with respect to x.

I.e.

Find

\displaystyle \frac{d}{dx}\ \left(\ \int_3^x sin(t^{5}) \, dt\ \right)\ .​

 

[LearninDaMath]

The limits od the integral are 3 and x, so the integral is a function of x, not t.

The derivative of this integral is with respect to x.

I.e.

Find

\displaystyle \frac{d}{dx}\ \left(\ \int_3^x sin(t^{5}) \, dt\ \right)\ .​

Thanks, I noticed that right after I submitted the post and editted with the correct notation right afterwards.

 

[zjmarlow]

Is this just 5x^4 sin(x^{5}) ?

 

[SammyS]

Is this just 5x^4 sin(x^{5}) ?

No.

The anti-derivative of 5x^4 \sin(x^{5}) is -\cos(x^{5}) +C,

not \displaystyle \int \sin(x^{5}) dx\ .

 

[zjmarlow]

So this should be sin(x^{5}), correct?

 

[HallsofIvy]

No, it isn't. What is the derivative of sin(u(x)) with respect to x. What is the derivative of cos(u(x)) with respect to x?

 

[zjmarlow]

cos(u(x)) u'(x) and -sin(u(x)) u'(x).

 

[Chestermiller]

So this should be sin(x^{5}), correct?

This result looks correct to me. There is no x under the integral sign. I don't understand what Halls of Ivy is saying.

Chet

 

[SammyS]

So this should be sin(x^{5}), correct?

If you mean:

Is \displaystyle <br /> \frac{d}{dx}\ \left(\ \int_3^x sin(t^{5}) \, dt\ \right)=\sin(x^5)\ ?​

Then I agree with Chestermiller that, "Yes it is."

Here's how I arrive at that:

Let F(t) be an anti-derivative of sin(t⁵). In other words, let \displaystyle F(t)=\int sin(t^{5}) \, dt\ .

Then \displaystyle \int_3^x sin(t^{5}) \, dt=F(x)-F(3)\ .

Therefore, \displaystyle <br /> \frac{d}{dx}\ \left(\ \int_3^x sin(t^{5}) \, dt\ \right)=\frac{d}{dx}\left(F(x)-F(3)\right)=\sin(x^5)-0\ .

 

[Mark44]

oh, it says find the derivative...lol. I just saw the integral sign and began trying to integrate. I didn't even notice that there is a d/dt notation to the right of the integral clear as day.

"When all else fails, read (or reread) the instructions."