Author Topic: Integration (Read 3547 times) Tweet Share

d0minicz · « **Reply #15 on:** June 17, 2009, 09:26:58 pm »

Evaluate: $\int sec^4(\frac {x}{2}) tan^2(\frac {x}{2})dx$
thanks

kamil9876 · « **Reply #16 on:** June 17, 2009, 09:37:41 pm »

$=\int sec^2\frac{x}{2}{(tan^2 (\frac{x}{2})+1)tan^2\frac{x}{2} dx$

let $u=tan(\frac{x}{2})$ and notice the $sec^2\frac{x}{2}$ on the side being the derivative

d0minicz · « **Reply #17 on:** June 17, 2009, 09:41:53 pm »

ah thanks ; shit i used 1- tan^x instead of 1+tan^2x lol
another two:
Evaluate: $\int sin(2x) \sqrt {1-sin(x)}dx$
and
$\int^{2\pi}_0 {\sqrt {1-cos(x)}}dx$
thanks alot

Mao · « **Reply #18 on:** June 17, 2009, 09:49:38 pm »

tough ones:

1.

$I = \int 2\sin(x)\cos(x)\sqrt{1-\sin(x)}\; dx$ , make the substitution $u=1-\sin(x)$

$\implies I = -2 \int (1-u)u^{-1/2}\; dv$ , the rest is trivial

2.

$1-\cos(x) = 2 \cdot \left( \frac{1 - \cos \left( 2\frac{x}{2}\right)}{2}\right) = 2\sin^2 \left( \frac{x}{2}\right)$

kamil9876 · « **Reply #19 on:** June 17, 2009, 09:55:18 pm »

first one:

$=\int 2sinxcosx\sqrt{1-sinx}dx$

let $u=1-sinx$
$sinx=1-u$
$\frac{du}{dx}=-cosx$

and so sub all that in to get:

$-2 \int \frac{du}{dx}(1-u)\sqrt{u} dx$

d0minicz · « **Reply #20 on:** June 17, 2009, 10:05:09 pm »

thanks
how would i convert this into a useful form:
$\int sin^2 (\frac {x}{2}) cos^2 (\frac {x}{2}) dx$

TrueTears · « **Reply #21 on:** June 17, 2009, 10:06:54 pm »

$sin(2a) = 2sinacosa$

sub in $a = \frac{x}{2}$

and square both sides

$sin(x) = 2sin(\frac{x}{2})cos(\frac{x}{2})$

$sin^2(x) = 4sin^2(\frac{x}{2})cos^2(\frac{x}{2})$

d0minicz · « **Reply #22 on:** June 19, 2009, 09:21:09 pm »

Find the area of the region enclosed by:
$y= sin2x , y=cosx ; \frac {-\pi}{2} <_ x <_ \frac {\pi}{2}$
thankkkkkkks
thats including $\frac {-\pi}{2}$ and $\frac {\pi}{2}$ dunno how to use latex fo that

TrueTears · « **Reply #23 on:** June 19, 2009, 09:28:13 pm »

First do a rough sketch.

Then find intersection points (Which I'm sure you know how to do)

This yields: $x = \frac{-\pi}{2}, \frac{\pi}{6} , \frac{\pi}{2}$

between $x = \frac{-\pi}{2}$ and $\frac{\pi}{6}$ , we have the integral $\int_{\frac{-\pi}{2}}^{\frac{\pi}{6}} cos(x) - sin(2x) dx$

Then from $x = \frac{\pi}{6} to \frac{\pi}{2}$ , we have the integral $\int_{\frac{\pi}{6}}^{\frac{\pi}{2}} sin(2x) - cos(x) dx$

Add the 2 integral results together and that should give the area required.

EDIT: \le and \ge for your latex query.

d0minicz · « **Reply #24 on:** June 19, 2009, 09:46:18 pm »

Another ....
same thing but for
$y= \sqrt {x} , y=6-x , y=1$
thanks..

TrueTears · « **Reply #25 on:** June 19, 2009, 10:01:31 pm »

Sketch the graphs.

Find the intersection between $y = \sqrt{x}$ and y = 1, which is x = 1

Then find intersection between y = 6 - x and $y = \sqrt{x}$ which is x = 4

and between y = 6 - x and y = 1 which is x = 5

We require this integral: $\int_1^4 \sqrt{x} - 1 dx + \int_4^5 6 - x - 1 dx$

d0minicz · « **Reply #26 on:** June 19, 2009, 10:56:35 pm »

jjsut a quick question with the above
how do you know its $\sqrt {x} - 1$ and not the other way around ?
thanks

TrueTears · « **Reply #27 on:** June 19, 2009, 10:58:14 pm »

Because $\sqrt{x}$ is above y = 1

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