Question

Trigonometric Substitution

Answer 1

\[\int\limits_{}^{}\frac{ x^2 }{ \sqrt{9-x^2} }\]

Answer 2

\[\int\limits_{}^{}\frac{ 9\sin^2(x) }{ 3\cos(x) }\]

Answer 3

Wait thats not right

Answer 4

\[\int\limits_{}^{}\frac{ 9\sin^2(x)3\cos(x) }{ 3\cos(x) }\]

Answer 5

Yup, correct.

Answer 6

Where do I go from here?
\[9\int\limits_{}^{}\sin^2(x)\]
?

Answer 7

Also it may help you to use something other than \(x\), just to indicate that they're different variables.

Do you know how to integrate \(\sin^2 x\)?

Answer 8

rewrite it as (1-cos^2) and split the integral is my first guess

Answer 9

Nah not entirely.

Answer 10

wait that wouldnt work lol

Answer 11

I mean yeah if you know how to integrate \(cos^2 x\) then that's fine.

Answer 12

u-sub?

Answer 13

Nope, just a straight up trig identity.\[\cos(2x) = 1 - 2 \sin^2 x\]

Answer 14

oh yea, double angle. Im a little confused how it helps though

Answer 15

OK, the first thing you do is write \(\sin^2 x\) as the subject of the equation.

Answer 16

yup

Answer 17

What did you come up with?

Answer 18

\[9\int\limits_{}^{}\sin^2(\theta)d \theta\]

Answer 19

Changed x to theta to make things less complicated

Answer 20

Yeah OK, now use that double-angle identity to write \(\sin^2 \theta\) in terms of \(\cos 2 \theta\)

Answer 21

\[\frac{ 9 }{ 2 }\cos(2\theta)=-\frac{ 1 }{ 2 }\]

Did I do that correctly?

Answer 22

I dont think I did, but Im not sure what I did wrong

Answer 23

\[\sin^2\theta \neq \cos(2\Theta)\]

Answer 24

\[\large\rm \sin^2\theta=\frac12(1-\cos2\theta)\]

Answer 25

oh duh, thanks @zepdrix