1. ## Integrations

1. The curve C has parametric equation x = sin t, y = sin 2t, 0 ≤ t ≤ pi/2.

If the region is revolved through $2 \pi$ radians about the x-axis, find the volume of the solid formed.
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So Volume = $\pi \int y^2 dx = \pi \int y^2 \frac {dx}{dt}dt$

$= \pi \int_{0}^{\frac{\pi}{2}} sin^2 2t \,cos\,t\, dt$

Then how could I integrate this?
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2. Find a general solution of the following differential equation.

$e^{x+y} \frac{dy}{dx} = x(2 + e^y)$
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So far I did,

$\int \frac{e^y}{2+e^y} dy = \int \frac{x}{e^x} dx$

$\int (1 - \frac{2}{2 + e^y})dy = \int x\, e^{-1}dx$

$y - 2 \,ln|2+e^y| = -xe^{-x} - e^{-x} + C$

But this is not right ans. The right ans is $ln|2+e^y| = -xe^{-x} - e^{-x} + C$

So where did I wrong?
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2. Originally Posted by geton
1. The curve C has parametric equation x = sin t, y = sin 2t, 0 ≤ t ≤ pi/2.

If the region is revolved through $2 \pi$ radians about the x-axis, find the volume of the solid formed.
-----------------------
So Volume = $\pi \int y^2 dx = \pi \int y^2 \frac {dx}{dt}dt$

$= \pi \int_{0}^{\frac{\pi}{2}} sin^2 2t \,cos\,t\, dt$

Then how could I integrate this?
-----------------------

2. Find a general solution of the following differential equation.

$e^{x+y} \frac{dy}{dx} = x(2 + e^y)$
-----------------------

So far I did,

$\int \frac{e^y}{2+e^y} dy = \int \frac{x}{e^x} dx$

$\int (1 - \frac{2}{2 + e^y})dy = \int x\, e^{-1}dx$

$y - 2 \,ln|2+e^y| = -xe^{-x} - e^{-x} + C$

But this is not right ans. The right ans is $ln|2+e^y| = -xe^{-x} - e^{-x} + C$

So where did I wrong?
-----------------------
Q1 Expand

$sin^2 2t \,cos\,t\, = 4 sin^2t \cos^3 t$ and use a substitution $u = cos\,t\,$

Q2

$\int \frac{e^y}{2+e^y} dy \ne y - 2 \,ln|2+e^y|$

If $u = 2+e^y$ then $du = e^y\, dy$ so $\int \frac{e^y}{2+e^y} dy = \int \frac{du}{u}$

which easily interates giving

$\int \frac{e^y}{2+e^y} dy = \,ln|2+e^y| + c$

3. Originally Posted by danny arrigo
Q1 Expand

$sin^2 2t \,cos\,t\, = 4 sin^2t \cos^3 t$ and use a substitution $u = cos\,t\,$

Q2

$\int \frac{e^y}{2+e^y} dy \ne y - 2 \,ln|2+e^y|$

If $u = 2+e^y$ then $du = e^y\, dy$ so $\int \frac{e^y}{2+e^y} dy = \int \frac{du}{u}$

which easily interates giving

$\int \frac{e^y}{2+e^y} dy = \,ln|2+e^y| + c$

Forgive my ignorance but how did you guys seperate the $e^{x+y}$ term?

Thanks

4. Originally Posted by tsal15
Forgive my ignorance but how did you guys seperate the $e^{x+y}$ term?

Thanks
One of the properties of exponents

$e^{x+y} = e^x \, e^y$

5. Originally Posted by geton

2. Find a general solution of the following differential equation.

$e^{x+y} \frac{dy}{dx} = x(2 + e^y)$
As you noted \begin{aligned}e^{x+y}\frac{dy}{dx}=x(2+e^y)&\impl ies \frac{e^y}{2+e^y}dy=\frac{x}{e^x}dx\\
&\implies\int\frac{e^y}{2+e^y}dy=\int\frac{x}{e^x} dx\\
&\implies \color{red}{\ln(2+e^y)}=-\left(\frac{x}{e^x}+\frac{1}{e^x}\right)+C\\
&\implies y=\ln\left(\exp\left(-\left\{\frac{x}{e^x}+\frac{1}{e^x}\right\}+C\right )-2\right)
\end{aligned}

The problem is that if you let $\varphi\mapsto 2+e^y$ in your integral you get $d\varphi=e^y~dy$ so you should have $\int\frac{e^y}{2+e^y}dy\stackrel{\varphi=2+e^y}{\l ongmapsto}\int\frac{d\varphi}{\varphi}$

6. Originally Posted by danny arrigo
Q1 Expand

$sin^2 2t \,cos\,t\, = 4 sin^2t \cos^3 t$ and use a substitution $u = cos\,t\,$

Ok,

Let $u = cos\, t$
$\frac{du}{dt} = - sin\,t$

Then?

Could you explain how to do this?

7. Originally Posted by geton
Ok,

Let $u = cos\, t$
$\frac{du}{dt} = - sin\,t$

Then?

Could you explain how to do this?
So far I thought

$\int 4 \, sin^2t \,cos^3 t \,dt$
$= - 4 \int u^3 \sqrt{1-u^2}\,du$

Is it the right way? Then?

8. Originally Posted by geton
So far I thought

∫4 sin^2 t cos^3 t dt
= 4∫ (1-cos^2 t)cos^3 t dt
= 4∫ (cos^3 t - cos^5 t) dt

Is it the right way? Then?
No.

First you note that $4 \sin^2 t \cos^3 t = 4 \sin^2 t \cos^2 t \cos t = 4 \sin^2 t (1 - \sin^2 t) \cos t$.

Then you make the substitution ${\color{red}u = \sin t}$ (NOT $u = \cos t$ which was no doubt a typo by danny). Note that $dt = \frac{du}{\cos t}$.

9. Originally Posted by Mathstud28
As you noted \begin{aligned}e^{x+y}\frac{dy}{dx}=x(2+e^y)&\impl ies \frac{e^y}{2+e^y}dy=\frac{x}{e^x}dx\\
&\implies\int\frac{e^y}{2+e^y}dy=\int\frac{x}{e^x} dx\\
&\implies \color{red}{\ln(2+e^y)}=-\left(\frac{x}{e^x}+\frac{1}{e^x}\right)+C\\
&\implies y=\ln\left(\exp\left(-\left\{\frac{x}{e^x}+\frac{1}{e^x}\right\}+C\right )-2\right)
\end{aligned}

The problem is that if you let $\varphi\mapsto 2+e^y$ in your integral you get $d\varphi=e^y~dy$ so you should have $\int\frac{e^y}{2+e^y}dy\stackrel{\varphi=2+e^y}{\l ongmapsto}\int\frac{d\varphi}{\varphi}$
My b and b.

There's not really problem.

The result given by danny is correct. What you've done (which is fine by the way) is to make explicit how the integration is done.

10. Originally Posted by mr fantastic
No.

First you note that $4 \sin^2 t \cos^3 t = 4 \sin^2 t \cos^2 t \cos t = 4 \sin^2 t (1 - \sin^2 t) \cos t$.

Then you make the substitution ${\color{red}u = \sin t}$ (NOT $u = \sin t$ which was no doubt a typo by danny). Note that $dt = \frac{du}{\cos t}$.
Thanks.

11. Originally Posted by mr fantastic
My b and b.

There's not really problem.

The result given by danny is correct. What you've done (which is fine by the way) is to make explicit how the integration is done.
Yes. I am sorry. I missed Danny's post.

12. Originally Posted by mr fantastic
No.

First you note that $4 \sin^2 t \cos^3 t = 4 \sin^2 t \cos^2 t \cos t = 4 \sin^2 t (1 - \sin^2 t) \cos t$.

Then you make the substitution ${\color{red}u = \sin t}$ (NOT $u = \cos t$ which was no doubt a typo by danny). Note that $dt = \frac{du}{\cos t}$.
Yep - a typo on my part. It should be $u = \sin t$