Question

Help Evaluating an Integral:

Answer 1

\[\huge \int\limits_{1}^{\sqrt{3}}\sqrt{4-x^2}-1 dx\]

Answer 2

\(x=2\sin(\theta)\)

Answer 3

why?

Answer 4

Because it helps solving the problem

Answer 5

you are allowed to just make a substitution like that?

Answer 6

oh i see. How would that now affect the upper/lower limits

Answer 7

Why not?
\[
\int\sqrt{4-x^2}-1\,dx\\
x=2\sin(\theta)\\
dx=2\cos(\theta)\,d\theta\\
\int\left(\sqrt{4-4\sin^2(\theta)}-1\right)2\cos(\theta)\,d\theta
\]

Answer 8

So that then becomes \[\int\limits_{}^{}(2\cos(\theta)-1)2\cos(\theta)d \theta\]

Answer 9

now what should i do

Answer 10

cos^2 x change into cos 2x and integrate it also change the limits due to substitution

Answer 11

cos 2x= 2 cos^2 x - 1

Answer 12

how should i change the limits

Answer 13

because in Q variable is x now is theta.

Answer 14

x = 2 sin a
1 =2 sin a
sin a =1/2
a= 30 degree
and when sqr 3 = 2 sin a
change it

Answer 15

Is this correct: \[\int\limits_{\pi/6}^{\pi/3}4\cos^2(\theta)-2\cos(\theta)d \theta\]

Answer 16

yes

Answer 17

i dont think it is.. i think i went wrong somewhere

Answer 18

no its fine. solve it

Answer 19

\[4\int\limits_{\pi/6}^{\pi/3}\cos^2(\theta)d \theta - 2\int\limits_{\pi/6}^{\pi/3}\cos(\theta)d \theta\]

Answer 20

\[
\cos(2\theta)=2\cos^2(\theta)-1
\]

Answer 21

that doesnt show up though

Answer 22

i need help with the integral of \[\cos^2(\theta)\]

Answer 23

so far i have \[4\int\limits_{\pi/6}^{\pi/3}\cos^2(\theta) d \theta + (-2\sin(\frac{\pi}{3})-(-2\sin(\frac{\pi}{6}))\]

Answer 24

\[
\cos^2(\theta)=\frac{\cos(2\theta)+1}{2}
\]
Does that make more sense now?

Answer 25

if you wants to finf cos^2 t direct use product formula otherwise change to cos 2t

Answer 26

which is \[4\int\limits_{\pi/6}^{\pi/3}\cos^2(\theta) d \theta +(1-\sqrt{3})\]

Answer 27

what is product formula

Answer 28

use cos 2 t instant of product formula its make it complected for u.

Answer 29

ok, so once i plug that in then what?

Answer 30

\[4\int\limits_{\pi/6}^{\pi/3}\frac{\cos(2\theta)+1}{2} d \theta\]

Answer 31

yes integrate it remember integration of cos ax =1/a (sin ax)

Answer 32

Does that thing become \[2\int\limits_{\pi/6}^{\pi/3}\cos(2\theta)+1 d \theta\]

Answer 33

so: \[2\int\limits_{\pi/6}^{\pi/3}\cos(2\theta) d \theta+2\int\limits_{\pi/6}^{\pi/3}1 d \theta\]

Answer 34

yes integrate it

Answer 35

and:\[2\int\limits_{\pi/6}^{\pi/3}\frac{1}{2}d \theta+2\int\limits_{\pi/6}^{\pi/3}\sin(2\theta)d \theta+2\int\limits_{\pi/6}^{\pi/3}1 d \theta\]

Answer 36

wait can i just use this: \[\int\limits_{}^{}\cos^2(x) dx=\frac{1}{2}(x+\sin(x)\cos(x))+C\]

Answer 37

ok so its \[\frac{\pi}{3}+(1-\sqrt{3})=\frac{\pi+3-3\sqrt{3}}{3}\]

Answer 38

woah nice thanks!

Answer 39

\[2\int\limits_{\pi/6}^{\pi/3}\cos(2\theta)\,d \theta+2\int\limits_{\pi/6}^{\pi/3}1 \,d \theta=[\sin(2\theta)+2\theta]_{\frac{\pi}{6}}^\frac{\pi}{3}\]

Answer 40

so thats where it comes from.. Ok

Answer 41

Did you integrated something twice?

Answer 42

yea i made a mistake and plugged the integral back into the operator