Question

integration the following:

Answer 1

\[\int\limits_{1}^{2} x^2 \div \sqrt{4-x^2}dx\]

Answer 2

Use
\(x=asin \theta\)

Answer 3

sorry it'snot the right one :D hold a secound i will get the question foryou

Answer 4

\[\int\limits_{1}^{2} 1\div \sqrt{1+\sqrt{x}}\]

Answer 5

i use x= tantheta

Answer 6

\[\Large \int\limits \frac{1}{\sqrt{1+\sqrt{x}}} \, dx?\]

Answer 7

Just u-sub 2 times

Answer 8

\[u=\sqrt{x} \\ \\ du=\frac{1}{2\sqrt x}\]
\[2 \int\limits \frac{u}{\sqrt{u+1}} \, du\]

Answer 9

Then s=u+1
ds=du
\[2 \int\limits \frac{s-1}{\sqrt{s}} \, ds\]
split denominator and integrate

Answer 10

snx

Answer 11

\[\int\limits_{2}^{1} tanx \div sinx -tanx\]

Answer 12

\[\int\limits (\frac{t}{s}-t)dx?\]

Answer 13

no :t/(s-t)

Answer 14

\[\int\limits \frac{ \tan x}{ \sin x - \tan x } \, dx\]

Answer 15

Hmm do you know weierstrass substitution?

Answer 16

http://en.wikipedia.org/wiki/Weierstrass_substitution

Answer 17

yes

Answer 18

i mult by cotane then by conjegate

Answer 19

Did you get something like
\[\LARGE \int\limits -\frac{4 u}{\left(u^2-1\right) \left(u^2+1\right) \left(\frac{2 u}{u^2-1}+\frac{2 u}{u^2+1}\right)} \, du\]

Answer 20

nope

Answer 21

i get c+1(1/2-sin^2x)^-1

Answer 22

weierstrass substitution is let
\[u=\tan(\frac{x}{2})\]
\[du=\frac{1}{2}\sec^2(\frac{x}{2})dx\]
-----------------------------------
then let
\[ \sin x=\frac{2 u}{u^2+1} ,~~~~~ \cos x=\frac{1-u^2}{u^2+1},~~~~ dx=\frac{2 du}{u^2+1}\]

Answer 23

okay snx

Answer 24

sam wouldn't be easier for the first question to let u = 1+\[\sqrt{x}\]

Answer 25

to make x =(u-1)^2

Answer 26

so easily get integration of u then reutrn bk
'

Answer 27

you can do that as well