# Homework Help: Solving an integral using an special substitution

1. Jun 27, 2010

### Telemachus

1. The problem statement, all variables and given/known data
Well, the exercise asks me to solve the next integral using an adequate substitution.
$$\displaystyle\int_{}^{}\sqrt[ ]{4-x^2}dx$$

3. The attempt at a solution
$$\displaystyle\int_{}^{}\sqrt[ ]{4-x^2}dx$$

What I did was:

$$x=2\sin\theta$$

$$dx=2\cos\theta d\theta$$

So, then I get
$$\displaystyle\int_{}^{}\sqrt[ ]{4-x^2}dx=4\displaystyle\int_{}^{}\cos^2\theta d\theta=cos\theta\sin\theta+\displaystyle\int_{}^{}\sin^2\theta d\theta=cos\theta\sin\theta+\displaystyle\int_{}^{}(1-\cos^2\theta) d\theta=$$
$$=cos\theta\sin\theta+\displaystyle\int_{}^{}d\theta-\displaystyle\int_{}^{}\cos^2\theta$$

$$\Rightarrow{5\displaystyle\int_{}^{}\cos^2\theta d\theta=cos\theta\sin\theta+\theta}\Rightarrow{\displaystyle\int_{}^{}\cos^2\theta d\theta=\displaystyle\frac{1}{5}cos\theta\sin\theta+\displaystyle\frac{\theta}{5}}$$

$$\theta=\arcsin(\displaystyle\frac{x}{2})$$
$$\sin\theta=(\displaystyle\frac{x}{2})$$

$$\cos\theta=\sqrt[ ]{1-\sin^2\theta}$$

$$\cos\theta=\sqrt[ ]{1-\displaystyle\frac{x^2}{4}}$$

$$\displaystyle\int_{}^{}\sqrt[ ]{4-x^2}=\displaystyle\frac{x}{10}\sqrt[ ]{1-\displaystyle\frac{x^2}{4}}+\displaystyle\frac{\arcsin(\displaystyle\frac{x}{2})}{5}$$

I think I've made a mistake, but I don't know in which step. Perhaps when I've get back to the original variables, or when I've solved by parts the integral of the square of the cosine.

Bye there, and thanks for posting.

2. Jun 27, 2010

### LCKurtz

Right there, try using the identity

$$\cos^2\theta = \frac{1+\cos(2\theta)}{2}$$

3. Jun 28, 2010

### Telemachus

Thank you LCKurtz.

4. Jun 28, 2010

### Telemachus

Can somebody tell me if this is right?

I must solve:
$$\displaystyle\int_{}^{}\displaystyle\frac{dx}{\sqrt[ ]{9+x^2}}$$

And I used the substitutions:
$$x=3sinht$$
$$dx=3coshtdt$$

And the identity:

$$cosh^2t-sinh^2t=1\Rightarrow{\cosht=\sqrt[ ]{1+sinh^2t}}$$

$$\displaystyle\int_{}^{}\displaystyle\frac{dx}{\sqrt[ ]{9+x^2}}=\displaystyle\int_{}^{}\displaystyle\frac{3coshtdt}{\sqrt[ ]{3^2+(3sinht)^2}}=\displaystyle\int_{}^{}\displaystyle\frac{3coshtdt}{\sqrt[ ]{3^2(1+sinh^2t)}}=\displaystyle\int_{}^{}\displaystyle\frac{coshtdt}{cosht}=\displaystyle\int_{}^{}dt=t+C=argsh(\displaystyle\frac{x}{3})+C$$

Last edited: Jun 28, 2010
5. Jun 28, 2010

### Staff: Mentor

I don't see anything obviously wrong, but I would have used a different substitution, tan(theta) = x/3.

6. Jun 28, 2010

### Telemachus

Thanks Mark.

$$\displaystyle\int_{}^{}\displaystyle\frac{dt}{\sin(t)}$$

I've solved this using the table, with definition of cosecant: $$\displaystyle\int_{}^{}\displaystyle\frac{dt}{\sin(t)}=\ln|csc(t)-cot(t)|+C$$

This was the result of solving: $$\displaystyle\int_{}^{}\displaystyle\frac{dx}{x\sqrt[ ]{1-x^2}}=\ln|csc(arcsin(x))-cot(arcsin(x))|+C$$

But now...
$$\displaystyle\int_{}^{}\displaystyle\frac{dx}{x^2\sqrt[ ]{4-x^2}}$$

$$x=2\sin(t)$$
$$dx=2\cos(t)dt$$

$$\displaystyle\int_{}^{}\displaystyle\frac{dx}{x^2\sqrt[ ]{4-x^2}}=\displaystyle\int_{}^{}\displaystyle\frac{2\cos(t)dt}{4\sin^2(t)\sqrt[ ]{2^2-(2\sin(t)^2}}=\displaystyle\frac{1}{4}\displaystyle\int_{}^{}\displaystyle\frac{dt}{\sin^2(t)}}=\displaystyle\frac{1}{4}\displaystyle\int_{}^{}csc^2(t)dt$$

I thought of solving by parts, but...

$$u=csc(t)$$
$$du=-csc(t)cot(t)dt$$

$$dv=csc(t)dt$$
$$v=\ln|csc(t)-cot(t)|$$

7. Jun 28, 2010

### LCKurtz

Hint: What is the derivative of cot(t)?

Also Forum procedures suggest that you post new questions in a new thread.

8. Jun 28, 2010

### Telemachus

Thanks LCKurtz. Srry for posting here again, is just that is the same kind of problem, I didn't wanted to flood with topics of mine.

I've arrived to this solution:
$$\displaystyle\frac{1}{4}\displaystyle\int_{}^{}csc^2(t )dt=-\displaystyle\frac{1}{4}cot(arcsin(\displaystyle\frac{x}{2}))$$

9. Jun 28, 2010

### Hurkyl

Staff Emeritus
For the record, the method the opening poster used (integration by parts) is perfectly valid, and IMHO fairly clever. His problem there is that he forgot the 4 outside of the inetgral.

10. Jun 28, 2010

### Telemachus

Thanks, I've solved it both ways :)

Using that substitution drives faster to the solution.

11. Jun 30, 2010

### hunt_mat

Another way (and I believe that this is the classical way of looking at it)
$$I=\int\frac{dx}{\sin x}=\int\frac{\sin xdx}{\sin^{2}x}=\int\frac{\sin xdx}{1-\cos^{2}x}$$
Then split using partial fractions to obtain:
$$I=\frac{1}{2}\int\frac{\sin x}{1-\cos x}dx+\frac{1}{2}\int\frac{\sin x}{1+\cos x}dx$$
The integral may be easily computed to arrive at:
$$I=-\frac{1}{2}\log (1-\cos x)+\frac{1}{2}\log (1+\cos x)$$
Now with a bit of algebra which includes a bit of trig identities, we arrive at the answer:
$$I=\log |\cot x+cosec x|$$

Last edited: Jun 30, 2010
12. Jun 30, 2010

### Telemachus

Thank you hunt_mat. I always enjoy seeing the many perspectives that surges from facing a problem in this forum.

13. Jun 30, 2010

### Staff: Mentor

In case anyone has lost track, the work below refers to the integral in post #6, not the one in the OP.
A different approach is the following:
$$\int\frac{dx}{\sin x} = \int csc~x~dx = \int csc~x\frac{csc~x + cot~x}{csc~x + cot~x}dx$$
$$= \int \frac{csc^2~x + csc~x cot~x}{csc~x + cot~x}dx$$

Now you can use an ordinary substitution, u = csc(x) + cot(x), du = -csc2(x) - csc(x)cot(x)dx, to get -ln|csc(x) + cot(x)| + C.

14. Jun 30, 2010

### hunt_mat

Mine was easier. )

15. Jun 30, 2010

### Staff: Mentor

I don't think so. Mine involved multiplying by 1 and an ordinary substitution. Yours requires partial fraction decomposition.

16. Jun 30, 2010

### Hurkyl

Staff Emeritus
It's even easier to just substitute -ln|csc(x)+cot(x)|+C for the integral. :tongue:

17. Jul 1, 2010

### hunt_mat

The multiplication by 1 bit was inspired, I wouldn't have thought about doing that. All the inspired bit that I did was again multiply by 1 but this time it was much more intuitive. Partial fraction are something everyone should know, it's a simple technique that is easily applied.