Integrating Trig Functions: Solving $\int{\frac{sin^3x}{cos^2x}dx}$

[Mentallic]

Homework Statement

$$\int{\frac{sin^3x}{cos^2x}dx}$$

The Attempt at a Solution

I've tried converting the trigs into more useful forms, but didn't find anything helpful there.
Some help please?

 

[Cyosis]

Write the integrand in terms of a tangent then use the identity $1+\tan^2 x=\sec^2 x$. Can you take it from there?

Edit: Substituting $\sin^2 x=1-\cos^2 x$ is easier.

 

[lanedance]

hey Mentallic, have you tried cos^2(x) = 1-sin^2(x) and some subsititution from there?

 

[Mentallic]

Using $1+tan^2x=sec^2x$

$$\int{(sec^2x-1)(sinx)dx}$$

$$=\int{tanxsecx-sinx}dx$$

I get stuck here with the tanxsecx part...


with $cos^2x=1-sin^2x$

$$\int{\frac{sin^3x}{1-sin^2x}dx}$$

$$\int{\frac{-sinx(1-sin^2x)+sinx}{1-sin^2x}dx}$$

$$\int{-sinx+\frac{sinx}{1-sin^2x}dx}$$

I'm just heading in the same direction as before... I must be using these identities in the wrong way. Any other hints?

 

[Дьявол]

Ok, you are heading to the right direction:
$$\int{-sinx+\frac{sinx}{1-sin^2x}dx}=$$

$$=\int{-sinxdx}+\int{\frac{sinx}{1-sin^2x}dx}=$$

$$=cosx+\int{\frac{sinx}{cos^2x}dx}$$

I. Use the substitution method.

$t=cosx$, so that

$$dt=-sinx$$

or

use the method integration by parts to solve the remaining integral:

$$\int u\, dv=uv-\int v\, du.\!$$

$$u=sinx , dv=\frac{1}{cos^2x}dx$$Good luck.

 

[Cyosis]

Using $1+tan^2x=sec^2x$

$$\int{(sec^2x-1)(sinx)dx}$$

$$=\int{tanxsecx-sinx}dx$$

I get stuck here with the tanxsecx part...

This is correct, the rest is overly complicated and only makes things harder. If you knew your trig derivatives by heart you would instantly recognize the primitive.

$$\tan x \sec x=\frac{\sin x}{\cos^2 x}$$

You know that the derivative of the cosine is the negative sine. This makes it clear you want to make a substitution involving cos.

with $cos^2x=1-sin^2x$

$$\int{\frac{sin^3x}{1-sin^2x}dx}$$

$$\int{\frac{-sinx(1-sin^2x)+sinx}{1-sin^2x}dx}$$

$$\int{-sinx+\frac{sinx}{1-sin^2x}dx}$$

I'm just heading in the same direction as before... I must be using these identities in the wrong way. Any other hints?

That is a very poor identity to use since you complicate the denominator this way. I told you $\sin^2x=1-\cos^2x$ would be easier, because then you would get that term in the numerator which makes things easy to split.

 

[Mentallic]

Ahh I see what you were intending now. I was able to convert

$\int{\frac{sinx}{cos^2x}dx}$ into $-\int{u^{-2}}du$ where $u=cosx$

Thanks a lot for the help.

 

[Cyosis]

Yep that's it. Secondly I would advice you to also memorize the derivatives of the sec, csc and cot functions. It will make spotting primitives a lot easier. For example:

$$\frac{d \sec x}{dx}=\sec x \tan x$$

This would have allowed you to solve your integral almost instantly.

 

[Mentallic]

$$\frac{d \sec x}{dx}=\sec x \tan x$$

This would have allowed you to solve your integral almost instantly.

Oh yeah I suppose the best tool in integration is rididculous amounts of practice to remember all possible results I would be asked in an exam hehe.

 

[Cyosis]

Experience is certainly a great asset when it comes to finding primitives. To solve the more common integrals you should be proficient in:

Knowing basic derivatives
Partial fractions
Completing the square
Trigonometric substitutions
Hyperbolic substitutions

Knowing these together with the basic integral theorems, such as integration by parts, will allow you to handle all the integrals they will throw at you during an exam.

 

[Mentallic]

Thanks for your advice and help Cyosis