Question

∫ (t^2)/√(16-t^2)

Answer 1

\[∫ \frac{t^2}{\sqrt{16-t^2}}dt\]Let \(t = 4 \sin \theta\) , \(dt = 4 \cos \theta d\theta\)
So, the integral becomes
\[\int\frac{(4\sin\theta)^2}{\sqrt{16-(4\sin\theta)^2}}(4\cos\theta d\theta)\]
Can you integrate it now?

Answer 2

Not really……

Answer 3

Where are you stuck at?

Answer 4

I suck at integration with sin/cosine…

Answer 5

Can you show the step where you are stuck at?

Answer 6

I don't even know where to start from your formula

Answer 7

Simplify the denominator first, what do you get?

Answer 8

4√(1-sin()²)

Answer 9

What is 1-(sinθ)^2?

Answer 10

Cos²

Answer 11

I get 4tan²(), but how do I integrate it?

Answer 12

No no no...
1 - (sinθ)^2 = (cosθ)^2 <- correct
So, \[4\sqrt{(1-sin^2 \theta)} = 4\sqrt{cos^2\theta}=4cos\theta\]The integral now becomes
\[\int\frac{(4\sin\theta)^2}{4cos\theta}(4\cos\theta d\theta)\]That is\[\int\frac{(16\sin^2\theta)4\cos\theta}{4cos\theta} d\theta\]Simplify the fraction, what do you get?

Answer 13

16∫sin²(x) dx

Answer 14

Yes. Can you integrate this?

Answer 15

Sorry I can't

*facepalm

Answer 16

Useful identity:
cos(2θ) = 1 - 2(sinθ)^2
That is
(sinθ)^2 = [ 1 - cos(2θ) ] /2

Answer 17

And replace 2x with u where du=2, got it. Thanks

Answer 18

You're welcome. Be careful with the variable you are using though.