1. ## integration - urgent help please

2 times the integral from 0 to 1 of (2t-3t^2) *sqrt(1-9t^2) dt.

Thank you very much.

2. Originally Posted by kittycat
2 times the integral from 0 to 1 of (2t-3t^2) *sqrt(1-9t^2) dt.

Thank you very much.
$\int_0^1(2t - 3t^2)\sqrt{1 - 9t^2}~dt$

First let's get that radicand simplified a bit:
Let $x = 3t \implies dx = 3dt$
$\int_0^1(2t - 3t^2)\sqrt{1 - 9t^2}~dt = \int_0^3 \left [ 2 \left ( \frac{x}{3} \right ) - 3 \left ( \frac{x}{3} \right ) ^2 \right ] \sqrt{1 - x^2} ~ \frac{dx}{3}$

$= \frac{1}{9} \int_0^3 (2x - x^2) \sqrt{1 - x^2} ~ dx$

$= \frac{1}{9} \int_0^3 2x \sqrt{1 - x^2} ~ dx - \frac{1}{9} \int_0^3 x^2 \sqrt{1 - x^2} ~ dx$

The first integral may be done using a simple substitution. I'll let you deal with that. For the second integral:
$\int_0^3 x^2 \sqrt{1 - x^2} ~ dx$

Take a moment and look at the graph I've posted below. You can easily see that this integrand does not exist beyond x = 1. So
$\int_0^3 x^2 \sqrt{1 - x^2} ~ dx = \int_0^1 x^2 \sqrt{1 - x^2} ~ dx$ <-- We need to do this, else the substitution below is illegal.

Now Let $x = sin(\theta) \implies dx = cos( \theta) d \theta$
$\int_0^1 x^2 \sqrt{1 - x^2} ~ dx = \int_0^{\pi/2}sin^2(\theta) \sqrt{1 - sin^2(\theta) } ~cos(\theta) ~ d \theta$

$= \int_0^{\pi/2}sin^2(\theta) ~ cos^2(\theta) ~ d \theta$

I'll let you do this one also and then collect all the pieces.

-Dan

3. Hello, kittycat!

A variation on Dan's solution . . .

$2 \int^1_0\,(2t-3t^2)\cdot\sqrt{1-9t^2}\,dt$
Let: $3t \:= \:\sin\theta\quad\Rightarrow\quad dt \:=\:\frac{1}{3}\cos\theta\,d\theta$
. . and: . $\sqrt{1-9t^2} \:=\:\cos\theta$

Substitute: . $2\int\left[2\left(\frac{1}{3}\sin\theta\right) - 3\left(\frac{1}{3}\sin\theta\right)^2\right]\cdot\cos\theta \left(\frac{1}{3}\cos\theta\,d\theta\right)$

. . $= \;\;\frac{2}{9}\int(2\sin\theta - \sin^2\!\theta)\cos^2\!\theta\,d\theta \;\;=\;\;\frac{4}{9}\int\cos^2\!\theta\sin\theta\, d\theta - \frac{2}{9}\int\sin^2\!\theta\cos^2\!\theta\,d\the ta$

For the first integral, use: $u \,=\,\cos\theta$

For the second integral, we have:
. . $\frac{1}{4}\left(4\sin^2\!\theta\cos^2\!\theta\rig ht) \;=\;\frac{1}{4}\left(2\sin\theta\cos\theta\right) ^2 \;=\;\frac{1}{4}\!\cdot\!\sin^2\!2\theta \;=\;\frac{1}{8}(1 - \cos4\theta)$

Can you finish it?

4. Originally Posted by Soroban
Hello, kittycat!

A variation on Dan's solution . . .

Let: $3t \:= \:\sin\theta\quad\Rightarrow\quad dt \:=\:\frac{1}{3}\cos\theta\,d\theta$
. . and: . $\sqrt{1-9t^2} \:=\:\cos\theta$

Substitute: . $2\int\left[2\left(\frac{1}{3}\sin\theta\right) - 3\left(\frac{1}{3}\sin\theta\right)^2\right]\cdot\cos\theta \left(\frac{1}{3}\cos\theta\,d\theta\right)$
The only problem I have with this method is the integration limits on this last integral. I suppose you can restrict it at this point, as I did later on in my solution?

-Dan

5. what about for this integral,

the integral of t^2 *sqrt(1+9t^2) dt. How should I solve it? Thank you very much.

6. Originally Posted by kittycat

the integral of t^2 *sqrt(1+9t^2) dt. How should I solve it? Thank you very much.
Instead of using the sine function, use the tangent function. ( $1 + tan^2(x) = sec^2(x)$.)

-Dan

7. Quote:
Originally Posted by kittycat

the integral of t^2 *sqrt(1+9t^2) dt. How should I solve it? Thank you very much.

Instead of using the sine function, use the tangent function. (.)

-Dan

Hi Dan,

Could you please show me how to let the tangent function in this question? Sorry to ask you because I forgot the trig substitution. Thank you very much.

8. Originally Posted by kittycat
Quote:
Originally Posted by kittycat

the integral of t^2 *sqrt(1+9t^2) dt. How should I solve it? Thank you very much.

Instead of using the sine function, use the tangent function. (.)

-Dan

Hi Dan,

Could you please show me how to let the tangent function in this question? Sorry to ask you because I forgot the trig substitution. Thank you very much.
The same way that Soroban did, though in this case there are no restrictions due to the limits. So
$\int t^2\sqrt{1 + 9t^2}~dt$

Let $3t = tan(\theta) \implies 3 dt = sec^2(\theta)~d \theta$

So
$\int t^2\sqrt{1 + 9t^2}~dt = \int \frac{1}{9}tan^2(\theta) \sqrt{1 + tan^2(\theta)} ~ sec^2(\theta)~d\theta$

$= \frac{1}{9} \int tan^2(\theta) sec^3(\theta)~d\theta$

-Dan

9. ## help me again please

Hi,

I might be wrong in the process of calculation. But don't know why can't get the answer (given by the textbook) from your approach.

The answer given by the textbook is -11/108 *sqrt(10) -1/36*ln(sqrt(10)-3) -4/27

10. Originally Posted by kittycat

Hi,

I might be wrong in the process of calculation. But don't know why can't get the answer (given by the textbook) from your approach.

The answer given by the textbook is -11/108 *sqrt(10) -1/36*ln(sqrt(10)-3) -4/27

That's because your book isn't right. I get $\frac{32 - 3\pi}{216}$.

-Dan

11. he same way that Soroban did, though in this case there are no restrictions due to the limits. So

Let

So

========================
Referring to : the integral from 0 to 1 of t^2 *sqrt(1+9t^2) dt.

I got to the same answer as you posted. But how should i evaluate it if it is =>
the integral from 0 to 1 of t^2 *sqrt(1+9t^2) dt.

Could you please guide me to the final answer of this integration?

Thank you very much.