Solving an integral by substitution method

[Telemachus]

Homework Statement

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.

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

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.

 

[Office_Shredder]

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?

 

[Mark44]

I'm dealing with undefined integrals now.

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.

 

[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!

 

[malicx]

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!

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

 

[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:

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

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

 

[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.

 

[Telemachus]

Thank you gomunkul51.

 

[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?

 

[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?

 

[gomunkul51]

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

If you want to use substitution try:

t=1/sin^2(x)

differentiate it and see how it can help you :)

*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 !

 

[Telemachus]

This is what I did:

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

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

Then:

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

 

[gomunkul51]

On:

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

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

 

[Telemachus]

Thank you gomunkul51.

 

[hunt_mat]

One of the things you could have done is used
$$\cos^{3}x=\cos x (1-\sin^{2})<br /> [\tex]<br /> and then you can integrate easily.$$

 

[Telemachus]

Ok. I have one more doubt about this. For example in this case, as I solved:

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

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...

 

[hunt_mat]

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

 

[Mark44]

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

In other words, when n = -1.

 

[gomunkul51]

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

 

[hunt_mat]

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