2 times the integral from 0 to 1 of (2t-3t^2) *sqrt(1-9t^2) dt.
Thank you very much.
$\displaystyle \int_0^1(2t - 3t^2)\sqrt{1 - 9t^2}~dt$
First let's get that radicand simplified a bit:
Let $\displaystyle x = 3t \implies dx = 3dt$
$\displaystyle \int_0^1(2t - 3t^2)\sqrt{1 - 9t^2}~dt = \int_0^3 \left [ 2 \left ( \frac{x}{3} \right ) - 3 \left ( \frac{x}{3} \right ) ^2 \right ] \sqrt{1 - x^2} ~ \frac{dx}{3}$
$\displaystyle = \frac{1}{9} \int_0^3 (2x - x^2) \sqrt{1 - x^2} ~ dx$
$\displaystyle = \frac{1}{9} \int_0^3 2x \sqrt{1 - x^2} ~ dx - \frac{1}{9} \int_0^3 x^2 \sqrt{1 - x^2} ~ dx$
The first integral may be done using a simple substitution. I'll let you deal with that. For the second integral:
$\displaystyle \int_0^3 x^2 \sqrt{1 - x^2} ~ dx$
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
$\displaystyle \int_0^3 x^2 \sqrt{1 - x^2} ~ dx = \int_0^1 x^2 \sqrt{1 - x^2} ~ dx$ <-- We need to do this, else the substitution below is illegal.
Now Let $\displaystyle x = sin(\theta) \implies dx = cos( \theta) d \theta$
$\displaystyle \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 $
$\displaystyle = \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
Hello, kittycat!
A variation on Dan's solution . . .
Let: $\displaystyle 3t \:= \:\sin\theta\quad\Rightarrow\quad dt \:=\:\frac{1}{3}\cos\theta\,d\theta $$\displaystyle 2 \int^1_0\,(2t-3t^2)\cdot\sqrt{1-9t^2}\,dt$
. . and: .$\displaystyle \sqrt{1-9t^2} \:=\:\cos\theta$
Substitute: .$\displaystyle 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) $
. . $\displaystyle = \;\;\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: $\displaystyle u \,=\,\cos\theta$
For the second integral, we have:
. . $\displaystyle \frac{1}{4}\left(4\sin^2\!\theta\cos^2\!\theta\rig ht) \;=\;\frac{1}{4}\left(2\sin\theta\cos\theta\right) ^2 \;=\;\frac{1}{4}\!\cdot\!\sin^2\!2\theta \;=\;\frac{1}{8}(1 - \cos4\theta)$
Can you finish it?
Quote:
Originally Posted by kittycat
what about for this integral,
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
$\displaystyle \int t^2\sqrt{1 + 9t^2}~dt$
Let $\displaystyle 3t = tan(\theta) \implies 3 dt = sec^2(\theta)~d \theta$
So
$\displaystyle \int t^2\sqrt{1 + 9t^2}~dt = \int \frac{1}{9}tan^2(\theta) \sqrt{1 + tan^2(\theta)} ~ sec^2(\theta)~d\theta$
$\displaystyle = \frac{1}{9} \int tan^2(\theta) sec^3(\theta)~d\theta$
-Dan
he same way that Soroban did, though in this case there are no restrictions due to the limits. So
Let
So
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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.