HELP: Integration using Substitution (1 Viewer)

[hkgirl]

PLEASE HELP...
this is urgent & i seriously don't get this topic!!

1) By using the substitution
a) u = 2x + 7
b) u^2 = 2x + 7

find the integration of:

x square root (2x+7) dx



2) Use the substituion u = 5 - x
find the integration of:

x squareroot (5-x) dx



3) Using u = 1 + x^4

find the integration of

(x^3)/ (1+x^4)^5 with end points 1 and 0



THANKS in advance!!
will be much appreciated!

 

[SoulSearcher]

1) i) u = 2x + 7, therefore x = (u-7)/2
du/dx = 2, therefore du/2 = dx
Int. [x*rt(2x+7)] dx
= Int. [(u-7)/2 * rt(u)] * du/2
= 1/4 Int. [(u-7)rt(u)] du
= 1/4 Int. [u^3/2 - 7u^1/2] du

From there, you integrate as normal, then substitute in u = 2x+7 when you have finished integrating.

ii) u² = 2x + 7, therefore x = (u²-7)/2
dx/du = u, therefore dx = u du
u = +/- rt(2x+7), however we will deal with only the positive sign here.
Int. [x*rt(2x+7)] dx
= Int. [(u²-7)/2 * u]*u du
= 1/2 Int. [(u²-7)u²] du
= 1/2 Int. [u⁴ - 7u²]du

You can integrate normally from there, and substitute u² back in when appropriate.

2) u = 5-x, therefore x = 5-u
du/dx = -1, therefore dx = -du
Int. [x*rt(5-x)] dx
= Int. [(5-u)rt(u)]*-du
= Int. [(u-5)rt(u)] du
= Int. [u^3/2-5u^1/2] du

You can integrate normally from there.

3) x³/(1+x⁴)⁵ = 1/4 * 4x³/(1+x⁴)⁵
u = 1 + x⁴
du/dx = 4x³, therefore dx = du/4x³
x = 1, u = 2, similarly, x = 0, u = 1
Int. [x³/(1+x⁴)⁵] dx (1-->0)
= 1/4 Int. [4x³/(1+x⁴)⁵] dx (1-->0)
= 1/4 Int. [1/u⁵] du (2-->1)
= 1/4 [-1/4 * u^-4] (2-->1)
= -1/16 [1/u⁴] (2-->1)

Sub in your values of u, and you should get your answer that way.

 

[hkgirl]

omg thanks so much!!

you're sooo nice! you typed all that out for me!!

how did you type the powers??

 

[watatank]

x² = x[sup ] 2 [/sup ] without the spaces to write in superscript ie higher than normal.

x₂ = x[sub ] 2 [/sub ] without spaces to write in subscript (lower than normal)

 

[SoulSearcher]

omg thanks so much!!

you're sooo nice! you typed all that out for me!!

how did you type the powers??

It was absolutely no problem at all

Nice? Aww

 

[Fish Sauce]

I love you all so much, I was stuck on that exact same question. That EXACT one.

Stupid textbook with no examples for those type of question.

 

[bos1234]

how come u neevr wirte worked soln. like that for me?

 

[SoulSearcher]

Usually because I rarely have the time anymore to write out worked solutions.

 

[Fish Sauce]

Just a quick question.

I'm trying to solve
Int. sinxcosx dx

I swear I've done it right (and checks on the internet confirm my answer) but the book says that the answer is
-1/4 cos2x + c

I did it like this:
u = cosx
du = -sin dx

- Int. -cosxsinx dx
= - Int. u du
= -u^2 / 2
= - cos^2 x / 2
= - 1/2 cos^2x

Can anyone confirm my answer?
Thanks!

 

[jb_nc]

its right

 

[elseany]

theres two ways to do that integral

what the text book did was let sinxcosx = (sin2x)/2

so (1/2)*Int.sin2x dx = -1/4cos2x + c

they're both right, the difference is in the constant

 

[Mattamz]

You are both right

Note:
cos2x = (cosx)^2 - (sinx)^2
= 2(cosx)^2 -1

(cosx)^2 = (cos2x + 1)/2

From this we replace ur last line of working:
=((cosx)^2)/2 + C
=((cos2x+1)/2)/2 + C
=(cos2x)/4 +1/4 + C
= (cos2x)/4 + c , (since 1/4 is a constant)

Personally i would have used the subsitution:
u = sinx
du = cosx.dx
Not that it really matters

 

[Fish Sauce]

Ah, righto. Thanks all!