# U substitution on ∫sin(x^5)dx

## Homework Statement

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

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

## 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?

Related Calculus and Beyond Homework Help News on Phys.org
vela
Staff Emeritus
Homework Helper
You don't. Note that the problem says to use the fundamental theorem of calculus. Don't ignore that bit of info.

SammyS
Staff Emeritus
Homework Helper
Gold Member

## Homework Statement

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

## 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.

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
Staff Emeritus
Homework Helper
Gold Member
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
$\displaystyle \frac{d}{dx}\ \left(\ \int_3^x sin(t^{5}) \, dt\ \right)\ .$​

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.

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

SammyS
Staff Emeritus
Homework Helper
Gold Member
Is this just $5x^4 sin(x^{5})$ ?
No.

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

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

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

HallsofIvy
Homework Helper
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?

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

Chestermiller
Mentor
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
Staff Emeritus
Homework Helper
Gold Member
So this should be $sin(x^{5})$, correct?
If you mean:
Is $\displaystyle \frac{d}{dx}\ \left(\ \int_3^x sin(t^{5}) \, dt\ \right)=\sin(x^5)\ ?$​

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

Mark44
Mentor
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.