Question

Trig Substitution: Calculate the given integral.

Answer 1

\[\int\limits_{}^{}\frac{ 4x^2 }{ \sqrt(16-x^2) }dx\]

Answer 2

I know that x=4sin(theta) dx=4cos(theta) and sqrt(16-x^2)=4cos(theta)

Answer 3

Sounds good so far. =)

Answer 4

and I know the awnser is

\[-2x\sqrt(16-x^2)+32\sin^{-1} (x/4)+C\]

Answer 5

Now when I plug in

Answer 6

\[\int\limits_{}^{}\frac{ 4(4\sin(\theta))^2 }{ 4\cos(\theta) }4\cos(\theta)d\theta\]

Answer 7

So I cancel 4cos(theta)

\[\int\limits 64\sin^2(\theta)\]

Answer 8

oh my bad
\[\int\limits 64\sin^2(\theta)d\theta\]

Answer 9

How do I integrate this?

Answer 10

What I do is use the trig identity:

\[\frac{1}{2}-\frac{\cos(2 \theta)}{2}=\sin^2(\theta)\]

Answer 11

OH! so its just
\[64 \int\limits (\frac{1}{2})-\frac{\cos2\theta)}{2}\]

Answer 12

\[d \theta\]

Answer 13

oh my bad

Answer 14

haha all good.

Answer 15

I just got to excited that I forgot it

Answer 16

My appologies but,so far I got
\[\int\limits 32-32\cos(2\theta)\]
then
\[32x-16\sin(2\theta)+C\]

Answer 17

But I know thats wrong

Answer 18

What am I missing?

Answer 19

Well you have theta and you need it all in terms of x's. Use another trig identity for sin(2 theta)

Answer 20

XD oh my im not up today

Answer 21

\[2\sin \theta \cos \theta = \sin (2 \theta)\]

Answer 22

Should work out better now. =)

Answer 23

\[32\theta-16\sin(2\theta)\]

Answer 24

\[32\theta-32\sin(\theta)\cos(\theta)\]

Answer 25

So now I substitute back to x

Answer 26

\[\theta=\sin^{-1} (x/4)\]
\[4\cos(\theta)=\sqrt(16-x^2)\]

Answer 27

\[\cos(\theta)=\sqrt(16-x^2)/4\]

Answer 28

gahh still stumped

Answer 29

because it sums to

\[32\sin^{-1}(x/4)-8xsqrt(16-x^2)\]

Answer 30

which is not the right awnser