Question

i need help with trignometric integration
i cant type my question here

Answer 1

can you `try` to type it? :) I can format it for you.

Answer 2

\[\int\limits_{?}^{?} \sqrt1-x{2} dx\]

Answer 3

is everything suppose to be under the square root?

Answer 4

\[\large \int\limits \sqrt{1-x^2}dx\]

Answer 5

no just the 1-x square

Answer 6

\[\large \int\limits \sqrt1-x^2\;dx\]
Like that? D:

Answer 7

yes

Answer 8

the first one u typed was good

Answer 9

oh ok.

Answer 10

i mean 1-x square comes under the sqare root

Answer 11

Let's think back to our most basic trig identity.\[\large \sin^2\theta+\cos^2\theta=1\]If we subtract sine from each side we get,\[\large \cos^2\theta=1-\sin^2\theta\]


The reason we would want to make a trig sub, is to get rid of the addition/subtraction between the terms. By doing so, we'll be able to take the square root much easier.

Answer 12

So we currently have, under the root, \(\large 1-x^2\)
But we would like to have something like \(\large 1-\sin^2\theta\) so we can simplify it down using our identity.

Answer 13

\(\large 1-(\color{orangered}{x})^2\)

And we need to change it to,
\(\large 1-(\color{orangered}{\sin\theta})^2\)

Can you see a substitution that might work for us? `Hint: it's the orange part :D`

Answer 14

x

Answer 15

lol, yes XD it will involve x.. too confusing still? ok ok ok

Answer 16

We want to let \(\large x=\sin\theta\)

Which will change our integral to,\[\large \int\limits \sqrt{1-\sin^2\theta}\;dx\]

Answer 17

ok

Answer 18

now our integral has a function of `theta`, (a different variable than what we started with).
So we need the differential (dx) to be in terms of theta.

So we'll take the derivative of our substitution.
\[\large x=\sin \theta\]Taking the derivative with respect to theta gives us,\[\large \frac{dx}{d \theta}=\cos \theta\]

We'll rewrite the d(theta) on the right. Think of this process as multiplication for now.\[\large dx=\cos \theta \;d \theta\]

Answer 19

\[\large \color{royalblue}{dx=\cos \theta \;d \theta}\]We'll plug this in for our dx.

\[\large \int\limits\limits\limits \sqrt{1-\sin^2\theta}\;(\color{royalblue}{dx}) \qquad = \qquad \int\limits\limits\limits \sqrt{1-\sin^2\theta}\;(\color{royalblue}{\cos \theta d \theta})\]

Answer 20

It's a bit of a tricky process if you're not familiar with it D:
We have to replace not only the x's when we make a substitution. But also the dx.

Answer 21

Using our trig identity that we talked about earlier, we can simplify the part under the root.\[\large \int\limits\sqrt{\cos^2\theta}\;(\cos \theta d \theta)\]

Answer 22

yea ok

Answer 23

And then from here, you can easily take the square root. Giving you a bit of an easier problem.

Answer 24

can you help me with that too?
i was just wnna solve the complete problem

Answer 25

Solve it completely? Sure.
Maybe seeing the full process will help you understand it.

Answer 26

if u can show me once i can do the other problems by myself please

Answer 27

yea that willl help me.thanks

Answer 28

Taking the square root, the multiplying the cosines gives us,
\[\large \int\limits\limits\sqrt{\cos^2\theta}\;(\cos \theta d \theta) \qquad = \qquad \int\limits \cos^2\theta \;d \theta\]

Answer 29

ok

Answer 30

From here, we need to apply the `Half-Angle Formula for Cosine`:

\[\large \cos^2\theta=\frac{1}{2}(1+\cos2\theta)\]

Answer 31

It might look messier, but it actually helps us. It get's rid of the exponent on the cosine.

Answer 32

\[\large \int\limits\cos^2\theta\;d\theta \qquad = \qquad \frac{1}{2}\int\limits 1+\cos2\theta\;d\theta\]

Answer 33

And after all that big fussy mess, we can FINALLY do the integration step!

Answer 34

So what does 1 give you, when you integrate it?

Answer 35

x

Answer 36

Yes very good. Since our integral is in terms of theta, it will give us theta. :)\[\large \frac{1}{2}\int\limits\limits 1+\cos2\theta\;d\theta \qquad = \qquad \frac{1}{2}\left(\theta+?\right)+C\]

Answer 37

Do you know how to integrate the cosine term?

Answer 38

no

Answer 39

hmm you really should try to get comfortable with that.
It's going to make problems like this a lot harder if you can't handle steps like this.

I mean... it's a simple step to integrate it.
But if you're not familiar with the idea, you probably will need to do a `u substitution`. Making another substitution within this substitution is just a bit messy. :)

Answer 40

are we gonna get any answer or we still have lot to do?

Answer 41

Ok ok let's not do a substitution, let's just think about this a sec.

\[\large \int\limits \cos2\theta\;d \theta\]

What's the integral of cosine? sine right?

\[\large \int\limits\limits \cos2\theta\;d \theta \qquad = \qquad \sin2\theta\]But now something is happening with the 2 right? :O

Answer 42

Ummm we're fairly close :3

Answer 43

ok

Answer 44

If you took the derivative of \(\large \sin2\theta\) a multiple of 2 would pop out due to the chain rule.

When we integrate, we divide by that coefficient instead.\[\large \int\limits\limits\limits \cos2\theta\;d \theta \qquad = \qquad \frac{1}{2}\sin2\theta\]
(Multiplying by 1/2 is the same as dividing by 2)

Answer 45

So this is what that gives us,\[\large \frac{1}{2}\int\limits\limits\limits 1+\cos2\theta\;d\theta \qquad = \qquad \frac{1}{2}\left(\theta+\frac{1}{2}\sin2\theta\right)+C\]

Answer 46

And from here, we want to rewrite our answer in terms of `x`, since that's the way the problem was presented to us.

Answer 47

If we go back to our initial substitution, we want to solve for theta.\[\large x=\sin\theta \qquad \rightarrow \qquad \arcsin x=\theta\]

Answer 48

\[\large \frac{1}{2}\left(\arcsin x+\frac{1}{2}\sin2\theta\right)+C\]

Answer 49

For these types of problems, you REALLY need to be comfortable with your trig. There are so many little identities we end up using! :D

Answer 50

To deal with the second term.. we'll use the `Double Angle Formula for Sine`:\[\large \sin2\theta=2\sin \theta \cos \theta\]

So our problem becomes,\[\large \frac{1}{2}\left(\arcsin x+\frac{1}{2}\sin2\theta\right)+C \qquad = \qquad \frac{1}{2}\left(\arcsin x+\sin\theta\cos\theta\right)\]

Answer 51

And then from here.. ugh this part is going to confuse you I bet.. :3
As if this problem hasn't been hard enough.. huh?

Answer 52

yea

Answer 53

Using our original substitution, we'll draw a triangle.

\[\large x=\sin \theta \qquad \rightarrow \qquad \frac{x}{1}=\sin \theta\]

Answer 54

Understand how I labeled this?
Sine is Opposite over Hypotenuse

Answer 55

yea

Answer 56

Solving for the missing side of the triangle gives us,