Integration ...... again

**Beard** · December 29th 2008, 07:55 AM

Once again I am stumped by integration with substitution. I have two problems which are probably a breeze for some of you, but I just seem to be doing this --->

.

1. $\int\frac{4x}{x^2 +3}$ where the substitution is $u = x^2 + 3$

and

2. $\int\cos^3(x)$ where the substitution is $s = \sin(x)$

Thanks for any help

**Mush** · December 29th 2008, 07:59 AM

Originally Posted by Beard

Once again I am stumped by integration with substitution. I have two problems which are probably a breeze for some of you, but I just seem to be doing this --->

.

1. $\int\frac{4x}{x^2 +3}$ where the substitution is $u = x^2 + 3$

and

2. $\int\cos^3(x)$ where the substitution is $s = \sin(x)$

Thanks for any help

First things first, remember to put in your "dx", when writing the integral. That thing is crucial to subsitution.

1. $\int\frac{4x}{x^2 +3} dx$ where the substitution is $u = x^2 + 3$

and

2. $\int\cos^3(x)dx$ where the substitution is $s = \sin(x)$

1. $\frac{du}{dx} = 2x$

Hence $xdx = \frac{1}{2}du$ . Just replace xdx with that, and replace x^2+3 with u.

2. $\frac{ds}{dx} = cos(x)dx$

$ds = cos(x)dx$

$I = \int\cos^3(x)dx = \int cos^2(x).cos(x)dx$

You need to write $cos^2(x)$ in terms of $sin(x)$ so you can use your substitution, s.

**galactus** · December 29th 2008, 08:02 AM

[quote=Beard;241407]Once again I am stumped by integration with substitution. I have two problems which are probably a breeze for some of you, but I just seem to be doing this --->

.

1. $\int\frac{4x}{x^2 +3}$ where the substitution is $u = x^2 + 3$

If we make the sub $u=x^{2}+3, \;\ du=2xdx, \;\ \frac{du}{2}=xdx$ . We make the subs and get:

$2\int\frac{1}{u}du$

Now, go to town.

**Chop Suey** · December 29th 2008, 08:02 AM

Originally Posted by Beard

Once again I am stumped by integration with substitution. I have two problems which are probably a breeze for some of you, but I just seem to be doing this --->

.

1. $\int\frac{4x}{x^2 +3}$ where the substitution is $u = x^2 + 3$

and

2. $\int\cos^3(x)$ where the substitution is $s = \sin(x)$

Thanks for any help

Rewrite $\int\frac{4x}{x^2 +3}~dx$ as $2\int\frac{2x}{x^2 +3}~dx$

Sub $u = x^2 + 3 \implies du = 2x~dx$ . This switches the integral to:

$2\int\frac{1}{u}~du$

Which should be a standard form you should pick up right away.

2. Subbing $s = \sin{x} \implies ds = \cos{x}~dx$ switches the integral to:

$\int \cos^2{x}~ds$

Notice how we managed to cancel one factor of x. Using the identity $\cos^2{x} = 1-\sin^2{x}$ , the integral becomes:

$\int (1-s^2)~ds$

$\ldots$

**Rapha** · December 29th 2008, 08:04 AM

Mush gave you some very good hints

The solution for your first problem is:

$\int \frac{4x}{x^2+3} dx$

u:=x^2 + 3
=> u' = 2x

$u' = \frac{du}{dx} => dx = \frac{du}{u'}$

$\int \frac{4x}{x^2+3} dx = \int \frac{4x}{u}*\frac{du}{u'}$

$= \int \frac{4x}{u}*\frac{du}{2x} = = \int \frac{2}{u}*du$

I hope you know the answer now!

Originally Posted by Mush

First things first, remember to put in your "dx", when writing the integral. That thing is crucial to subsitution.

1. $\int\frac{4x}{x^2 +3} dx$ where the substitution is $u = x^2 + 3$

and

2. $\int\cos^3(x)dx$ where the substitution is $s = \sin(x)$

1. $\frac{du}{dx} = 2x$

Hence $xdx = \frac{1}{2}du$ . Just replace xdx with that, and replace x^2+3 with u.

2. $\frac{ds}{dx} = cos(x)dx$
$ds = cos(x)dx$

$I = \int\cos^3(x) = \int cos^2(x).cos(x)dx$

You can write cos^2(x) in terms of sin(x).

you should use:
sin^2+cos^2 = 1

**Beard** · December 29th 2008, 08:23 AM

Thanks but I had already done those processes with both of them, its just getting past that point that I find difficult.

i.e. the first one

$\frac{4x}{x^2 +3}$ and $u = x^2 +3$

then,as you were saying,

$\frac{du}{2} = x.dx$ which would then mean

$\frac{2}{u}$ . Carrying on, if I was then to integrate this I would get

$2\ln(x^2 +3) +c$ .
Is this correct?

then the second one I had done what you said before but I just get sort of tangled up.

$\cos^3(x).dx$

where $s = sin(x)$ due to the trigonometric identity

$\cos^2(x) = 1 -\sin^2(x)$ becomes

$(1-s^2)\cos(x)$ or $\cos - s^2\cos(x)$

also $\frac{ds}{dx} = cos(x)$ .

**Rapha** · December 29th 2008, 08:27 AM

Hey mate

Originally Posted by Beard

Thanks but I had already done those processes with both of them, its just getting past that point that I find difficult.

i.e. the first one

$\frac{4x}{x^2 +3}$ and $u = x^2 +3$

then,as you were saying,

$\frac{du}{2} = x.dx$ which would then mean

$\frac{2}{u}$ . Carrying on, if I was then to integrate this I would get

$2\ln(x^2 +3) +c$ .
Is this correct?

Yup, the first one is correct: 2 ln(x^2 +3) +c

Well done, Beard!

Originally Posted by Beard

then the second one I had done what you said before but I just get sort of tangled up.

$\cos^3(x).dx$

where $s = sin(x)$ due to the trigonometric identity

$\cos^2(x) = 1 -\sin^2(x)$ becomes

$(1-s^2)\cos(x)$ or $\cos - s^2\cos(x)$

also $\frac{ds}{dx} = cos(x)$ .

Well, it is

$\int cos^3 dx = \int cos^2(x) cos(x) dx$

s:= sin(x)
=> s' = cos(x)

$s' = \frac{ds}{dx} => dx = ds/s' = ds / cos(x)$

$\int cos^2(x) cos(x) dx = \int cos^2(x) *cos(x) \frac{ds}{cosx}$

$= \int cos^2(x) du = \int (1-sin^2(x) ) ds = \int (1-s^2) ds$

because s = sin(x)
Think about it!

**Beard** · December 29th 2008, 08:56 AM

And *ding* the lightbulb goes on, for the second one anyway. As it finishes:

$\int(1 - s^2).ds$ becoming $s-\frac{s^3}{3} +c$

Although you said I was right for the first question, my books disagrees, and although I would love to believe the book is wrong it generally isn't and it gives the answer of:

$-\frac{1}{(x^2 +3)^2} +c$

**Rapha** · December 29th 2008, 09:02 AM

Originally Posted by Beard

And *ding* the lightbulb goes on, for the second one anyway. As it finishes:

$\int(1 - s^2).ds$ becoming $s-\frac{s^3}{3} +c$

Although you said I was right for the first question, my books disagrees, and although I would love to believe the book is wrong it generally isn't and it gives the answer of:

$-\frac{1}{(x^2 +3)^2} +c$

It is
$\int\frac{4x}{x^2 +3} \not= -\frac{1}{(x^2 +3)^2} +c$

But why is that?

Because

$[-\frac{1}{(x^2 +3)^2} +c]' = \frac{4x}{(x^2 + 3)^3}$

Edit: $\int(1 - s^2).ds$ becoming $s-\frac{s^3}{3} +c$

Thats the right answer

**hatsoff** · December 29th 2008, 09:05 AM

Originally Posted by Rapha

It is
$\int\frac{4x}{x^2 +3} \not= -\frac{1}{(x^2 +3)^2} +c$

But why is that?

Because

$[-\frac{1}{(x^2 +3)^2} +c]' = \frac{4x}{(x^2 + 3)^3}$

I concur. Either the book is wrong, or else Mr. Beard has misread it.

**tom@ballooncalculus** · December 29th 2008, 10:51 AM

All settled, here, I know, but here's a couple of pics...

As usual, straight continuous lines diff/anti-diff with respect to x, straight dashed lines with respect to the dashed balloon expression.

Don't integrate - balloontegrate! Balloon Calculus: worked examples from past papers

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