# Math Help - intergrals

1. ## intergrals

The directions state: evaluate the indefinite integrals by using the given substitutions to reduces the integrals to standard form. There were two problems i seem to not get at all.
1. (integral) (9r^2dr/(sqrt(1-r^3)) u = 1-r^3 du = -3r^2
the part that threw me off was -3r^2 is in 9r^2. All of the other cases, i had to add something namely the reciprocal to the left of the integral and do not know what to do in this case.

At any rate, the answer came to be: -6(1-r^3)^(1/2) + c

2. (integral) (sqrt(x))(sin(x^(3/2)-1)^2dx u = x^(3/2)-1 du = (3/2)(x^(1/2))
2/3 (integral) (sin u)^2

2/3 * (-1/3)(sin u)^3 *(cos u) = (-2/9)(sin (x^(3/2)-1))^3 * cos (x^(3/2)-1).

The answer came out to be (1/3)(x^(3/2)-1)-(1/6)(sin(2x^(3/2)-2)) + c

2. Originally Posted by driver327
The directions state: evaluate the indefinite integrals by using the given substitutions to reduces the integrals to standard form. There were two problems i seem to not get at all.
1. (integral) (9r^2dr/(sqrt(1-r^3)) u = 1-r^3 du = -3r^2
the part that threw me off was -3r^2 is in 9r^2. All of the other cases, i had to add something namely the reciprocal to the left of the integral and do not know what to do in this case.

At any rate, the answer came to be: -6(1-r^3)^(1/2) + c
$\int\frac{9r^2}{\sqrt{1-r^3}}dr$

If $u=1-r^3$ then $du=-3r^2dr$. This means that $dr=-\frac{du}{3r^2}$ so the $r^2$ factor in the integral cancels, and the $9$ is divided by $3$. Do you see how this worked out? You're just substituting $dr=-\frac{du}{3r^2}$ into the integral.

$\int\frac{9r^2}{\sqrt{1-r^3}}dr=-\int\frac{3}{\sqrt{u}}du$

You should be able to evaluate that.

3. Originally Posted by driver327
The directions state: evaluate the indefinite integrals by using the given substitutions to reduces the integrals to standard form. There were two problems i seem to not get at all.
1. (integral) (9r^2dr/(sqrt(1-r^3)) u = 1-r^3 du = -3r^2
the part that threw me off was -3r^2 is in 9r^2. All of the other cases, i had to add something namely the reciprocal to the left of the integral and do not know what to do in this case.

At any rate, the answer came to be: -6(1-r^3)^(1/2) + c

2. (integral) (sqrt(x))(sin(x^(3/2)-1)^2dx u = x^(3/2)-1 du = (3/2)(x^(1/2))
2/3 (integral) (sin u)^2

2/3 * (-1/3)(sin u)^3 *(cos u) = (-2/9)(sin (x^(3/2)-1))^3 * cos (x^(3/2)-1).

The answer came out to be (1/3)(x^(3/2)-1)-(1/6)(sin(2x^(3/2)-2)) + c

$\int\frac{9r^2}{\sqrt{1-r^3}}\,dr=-3\int\frac{-3r^2dr}{\sqrt{1-r^3}}=-3\int\frac{1}{\sqrt{u}}\,du$ ...

$\int\sqrt{x}\sin^2\left(x^{3\slash 2}-1\right)dx=\frac{2}{3}\int\left[\left(\frac{3}{2}\sqrt{x}dx\right)\sin^2\!\left(x^ {3\slash 2}-1\right)dx\right]=\frac{2}{3}\int\sin^2u\,du$ ...

Tonio

4. I had to edit my original post. I forgot to write a negative sign. I hope that didn't confuse you. I'm sure you would have noticed anyway.

5. Originally Posted by tonio

$\int\frac{9r^2}{\sqrt{1-r^3}}\,dr=-3\int\frac{-3r^2dr}{\sqrt{1-r^3}}=-3\int\frac{1}{\sqrt{u}}\,du$ ...

$\int\sqrt{x}\sin^2\left(x^{3\slash 2}-1\right)dx=\frac{2}{3}\int\left[\left(\frac{3}{2}\sqrt{x}dx\right)\sin^2\!\left(x^ {3\slash 2}-1\right)dx\right]=\frac{2}{3}\int\sin^2u\,du$ ...

Tonio
on the second one, do you integrate just the exponent or both the exponent and the trig function?

6. Originally Posted by driver327
on the second one, do you integrate just the exponent or both the exponent and the trig function?

I hope I didn't understand you correctly, lest you're asking a rather weird thing: in $\int x^2dx$ do you "integrate just the exponent" or "both the exponent and the function x"? Absurd, isn't it?
The expression $\sin^2x$ is one single bundle, and it means the trigonometric function sine x raised to the second power: you can not "separate them! Now, if you want to know how to do it, here's an idea with some basic trigonometry:

$x=\int dx=\int (\cos^2x+\sin^2x)\,dx\,\Longrightarrow\,\int \sin^2x\,dx=x-\int \cos^2x\,dx=$ $x-\int(\cos2x+\sin^2x)\,dx$ .

Now just gather together stuff, solve the easy integral of $\cos2x$ and you're done.

Tonio

7. I thought a reverse chain rule (or something of that sort) had to be applied... but ok.