Find the integral of sinx/cos^3x dx

[tjbateh]

Homework Statement

Find the integral of sinx/cos^3x dx

Homework Equations

The Attempt at a Solution

How would I approach such a problem?

 

[snipez90]

If that's a cos(x) raised to the 3rd power in the denominator, then your integrand is just tan(x)sec^2(x) which is easy since the derivative of tan(x) is sec^2(x).

 

[tjbateh]

Wait so if I substitute U for cos(x)^3, then du=tan(x)sec^2(x)?

 

[Gregg]

No, he's saying that the integrand can be rewritten as tanxsec^2 x which can be easily integrated.

 

[tjbateh]

the integrand as in the whole problem? So sin(x)/cos(x)^3=tan(x)sec(x)^2??

 

[slider142]

the integrand as in the whole problem? So sin(x)/cos(x)^3=tan(x)sec(x)^2??

The integrand of an integral is the expression between the summa [tex]\int[/tex] and the differential dx. The equation you wrote above is correct for the integrand, which you can now integrate easily.

 

[Gregg]

the integrand as in the whole problem? So sin(x)/cos(x)^3=tan(x)sec(x)^2??

Yes. If

[tex]y=\sec^2(x)[/tex]


Then:


[tex]\frac{dy}{dx} = \cdots [/tex]

 

[tjbateh]

tan(x)sec(x)?? But i don't understand, what happens to the sin(x) in the numerator..It seems like were just talking about the denominator.

 

[slider142]

tan(x)sec(x)?? But i don't understand, what happens to the sin(x) in the numerator..It seems like were just talking about the denominator.

Do you remember the common definition of tan(x) = sin(x)/cos(x)?

 

[Gregg]

[tex] y=sec^2(x) \Rightarrow y=\frac{1}{cos(x)}\frac{1}{cos(x)}[/tex]

Use the product rule to differentiate that, you will see you have your derivative that is the [almost] the same as the integrand. Hence you have the answer.

 

[tjbateh]

wow, it finally makes sense! Thank you everyone!

 

[njama]

Here is much simpler approach:

[tex]\int{\frac{sinx}{cos^3x} dx}[/tex]

u=cos(x)

du=-sin(x)dx

dx=-du/sin(x)

[tex]\int{\frac{sin(x)}{u^3}*\frac{-du}{sin(x)}}=[/tex]

[tex]=-\int \frac{du}{u^3}[/tex]

 

[tjbateh]

This was more of the approach we learned in class. Would you then use the LN function?

 

[Bohrok]

You use [itex]\ln[/itex] if the integrand were 1/u where the denominator has a power of one, but for any other power, use the power rule for integrals.