# U substitution on ∫sin(x^5)dx

• LearninDaMath

## Homework Statement

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

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

## 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?

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

## 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.

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 $5x^4 sin(x^{5})$ ?

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?

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).

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

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

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.