# Solving an integral by substitution method

1. Jun 25, 2010

### Telemachus

1. The problem statement, all variables and given/known data
Hi there. I'm dealing with undefined integrals now. And I found this one that I don't know how to solve.
The problem statement says: Solve the next integrals using the substitution method.

$$\displaystyle\int_{}^{}\displaystyle\frac{\cos(x)}{\sin^3(x)}$$

3. The attempt at a solution
I've tried this way, but I don't know how to continue, and maybe there is a simpler way for solving it.

I thought of this substitution:

$$u=\sin^3(x)$$

$$du=3\sin^2(x)\cos(x)dx$$

$$du=3\cos(x)[1-cos^2(x)]dx$$
$$du=3\cos(x)dx-3\cos^3(x)dx$$

But then I don't know how to use it in my integral: $$\displaystyle\int_{}^{}\displaystyle\frac{\cos(x)}{\sin^3(x)}$$

Any suggestion?

Bye.

Last edited: Jun 25, 2010
2. Jun 25, 2010

### Office_Shredder

Staff Emeritus
Try a different substitution for u. I see two basic functions, one of which is the derivative of another one in your integral. Can you spot them?

3. Jun 26, 2010

### Staff: Mentor

I think you mean indefinite integrals, ones without limits of integration.

Also don't forget the dx. As the integration techniques become more complicated, omitting this will come back to bite you.

4. Jun 26, 2010

### Telemachus

Thanks. I've solved it using $$u=sin(x)$$

Sorry for the misspelling, my English is not too good :P

Now, here is another one, which I've also solved, but when I've tried to corroborate it on derive I've found a different result from which I've found:

I must solve: $$\displaystyle\int_{}^{}\displaystyle\frac{\cosh(\ln(x))}{3x}dx$$

So I've proceeded this way:

$$u=ln(x)$$
$$du=\displaystyle\frac{1}{x}dx$$

And then:

$$\displaystyle\int_{}^{}\displaystyle\frac{\cosh(\ln(x))}{3x}dx=\displaystyle\frac{1}{3}\displaystyle\int_{}^{}\displaystyle\frac{\cosh(\ln(x))}{x}dx=\displaystyle\frac{1}{3}\displaystyle\int_{}^{}\cosh(u)du=\displaystyle\frac{1}{3}sh(u)+C=\displaystyle\frac{1}{3}sh(ln(x))+C$$

But derive gives:

$$\displaystyle\int_{}^{}\displaystyle\frac{\cosh(\ln(x))}{3x}dx=\displaystyle\frac{x}{6}-\displaystyle\frac{1}{6x}$$

What am I doing wrong?

Bye there!

5. Jun 26, 2010

### malicx

i believe they are using the definitions of the cosh and sinh functions. $$\sinh x = \tfrac12\left(e^x - e^{-x}\right)$$

6. Jun 26, 2010

### Telemachus

Thanks, I think you're right.

Now, is this right?

$$\displaystyle\int_{}^{}\displaystyle\frac{\sin(t)\cos(t)}{\sqrt[ ]{3\sin(t)+5}}dt$$

I've used:

$$u=3\sin(t)+5\Rightarrow{\sin(t)=\displaystyle\frac{u-5}{3}}$$

$$du=3\cos(t)dt$$

Then:

$$\displaystyle\int_{}^{}\displaystyle\frac{\sin(t)\cos(t)}{\sqrt[ ]{3\sin(t)+5}}dt=\displaystyle\int_{}^{}\displaystyle\frac{\sin(t)3\cos(t)}{3\sqrt[2]{3\sin(t)+5}}dt=\displaystyle\int_{}^{}\displaystyle\frac{\displaystyle\frac{u-5}{3}du}{3\sqrt[2]{u}}=\displaystyle\int_{}^{}(u\sqrt[ ]{u}-5\sqrt[ ]{u})du=$$

$$=\displaystyle\int_{}^{}(u^{3/2}-5\sqrt[ ]{u})du=\displaystyle\frac{2}{5}u^{5/2}-\displaystyle\frac{10}{3}u^{3/2}+C=\displaystyle\frac{2}{5}(3\sin(t)+5)^{5/2}-\displaystyle\frac{10}{3}(3\sin(t)+5)^{3/2}+C$$

7. Jun 26, 2010

### gomunkul51

cosh(x) = (1/2)*(e^x + e^-x)
&
e^ln(x) = x
then:
cosh(ln(x)) = (1/2)*(e^ln(x) + e^-ln(x)) = (1/2)*(x - 1/x)

now you have no problem doing that integral.

Last edited: Jun 26, 2010
8. Jun 26, 2010

### Telemachus

Thank you gomunkul51.

9. Jun 26, 2010

### gomunkul51

For the Integral:

$$\displaystyle\int_{}^{}\displaystyle\frac{\cos(x)} {\sin^3(x)}$$

Try integration by parts.

V=1/sin(x)
U'=cos(x)/sin^2(x)

Can you do the rest?

10. Jun 26, 2010

### Telemachus

I've solved it using $$u=sin(x)$$ $$du=cos(x)$$

Now....

$$\displaystyle\int_{}^{}\cos^3(x)\sin^4(x)dx$$

$$u=\sin(x)$$
$$du=cos(x)dx$$

$$sin^2(x)+cos^2(x)=1\Rightarrow{cos^2(x)=1-\sin^2(x)}$$

$$\displaystyle\int_{}^{}\cos^3(x)\sin^4(x)dx=\displaystyle\int_{}^{}(1-\sin^2(x))\sin^4(x)\cos(x)dx=\displaystyle\int_{}^{}(\sin^4(x)-\sin^6(x))\cos(x)dx=\displaystyle\int_{}^{}(u^4-u^6)du=\displaystyle\frac{u^5}{5}-\displaystyle\frac{u^7}{7}+C=\displaystyle\frac{\sin^5(x)}{5}-\displaystyle\frac{\sin^7(x)}{7}+C$$

Did I make any mistake on this?

11. Jun 26, 2010

### gomunkul51

For:
$$\displaystyle\int_{}^{}\displaystyle\frac{\cos(x)} {\sin^3(x)}$$

If you want to use substitution try:

t=1/sin^2(x)

*for the future in those trig integral try to find your substitutions by picking not the whole nominator or denominator but just a part of it as in taking just sin^2(x) from sin^3(x) and integrating or differentiating a seeing where it got you !

12. Jun 26, 2010

### Telemachus

This is what I did:

$$\displaystyle\int_{}^{}\displaystyle\frac{\cos(x)} {\sin^3(x)}$$

$$u=\sin(x)$$
$$du=\cos(x)dx$$

Then:

$$\displaystyle\int_{}^{}\displaystyle\frac{\cos(x)}{\sin^3(x)}=\displaystyle\int_{}^{}\displaystyle\frac{du}{u^3}=\displaystyle\frac{u^{-2}}{-2}+C=\displaystyle\frac{1}{-2\sin^2(x)}+C$$

13. Jun 26, 2010

### gomunkul51

On:

$$\displaystyle\int_{}^{}\cos^3(x)\sin^4(x)dx$$

you right.
you did a great job, very efficient and quick integration.

14. Jun 26, 2010

### Telemachus

Thank you gomunkul51.

15. Jun 26, 2010

### hunt_mat

One of the things you could have done is used
$$\cos^{3}x=\cos x (1-\sin^{2}) [\tex] and then you can integrate easily. 16. Jun 26, 2010 ### Telemachus Ok. I have one more doubt about this. For example in this case, as I solved: [tex]\displaystyle\int_{}^{}\displaystyle\frac{\cos(x)} {\sin^3(x)}=\displaystyle\int_{}^{}\displaystyle\frac{du}{u^3}=\displaystyle\frac{u^{-2}}{-2}+C=\displaystyle\frac{1}{-2\sin^2(x)}+C$$

Particularly this part:
$$\displaystyle\int_{}^{}\displaystyle\frac{du}{u^3}=\displaystyle\frac{u^{-2}}{-2}+C$$
I just applied that $$\displaystyle\int_{}^{}u^ndu=\displaystyle\frac{u^{n+1}}{n+1}+C$$

right?

But when I have $$\displaystyle\int_{}^{}\displaystyle\frac{du}{u}=\ln(u)+C$$

I know that it is because of the that the derivative of $$ln(x)=\displaystyle\frac{1}{x}$$, and if I've used what I did in the previous case it would drive me into an undefined, I mean if I do

$$\displaystyle\int_{}^{}\displaystyle\frac{du}{u}=\displaystyle\int_{}^{}u^{-1}du=1/0$$ That if I use $$\displaystyle\int_{}^{}u^ndu=\displaystyle\frac{u^{n+1}}{n+1}+C$$

Well, I don't know if its clear what I'm trying to say, anyway....

17. Jun 26, 2010

### hunt_mat

The integral
$$\int x^{n}dx=\frac{x^{n+1}}{n+1}+c$$
doesn't apply with the case of 1/x

18. Jun 26, 2010

### Staff: Mentor

In other words, when n = -1.

19. Jun 27, 2010

### gomunkul51

You got it right, integral of 1/x is ln(x)
any other power of x you use that power rule you used.

20. Jun 27, 2010

### hunt_mat

It all comes down to the fundamental theorem of calculus really.