Thread: Techniques of integration: Trigonometric substitution

1. Techniques of integration: Trigonometric substitution

Hi,

I've solved this problem, but it's more of a nit-pick on something I tried previously in that according to the book was not correct, and now I'm just in need of someone's expertise on why I was not allowed to do this.

Anyway, the integral and solution is: -

$\displaystyle F(x) = \int sin^3(x)cos^3(x)\ dx = \frac{1}{48}cos^3(2x) - \frac{1}{16}cos(2x) + C$

In this case, I tried a substitution to begin with: -

$\displaystyle F(x) = \int sin^3(x)cos^3(x)\ dx = \int sin(x)[1 - cos^2(x)]cos^3(x)\ dx$

$\displaystyle Let\ u = cos(x),\ then\ du = - sin(x)\ dx$

$\displaystyle \Rightarrow F(u) = - \int (1 - u^2)u^3\ du = \int (u^5 - u^3)\ du = \frac{1}{6}u^6 - \frac{1}{4}u^4 + C$

$\displaystyle \Rightarrow F(x) = \frac{1}{6}cos^6(x) - \frac{1}{4}cos^4(x) + C$

Now, on the second attempt I made the appropriate double-angle substitution and achieved the solution I should have done. But, I can't see what is wrong with the former way I chose instead? Why can't I do it this way?

Regards

Ash

2. Re: Techniques of integration: Trigonometric substitution

This is not the answer but I think that your notation for $\displaystyle F(x)$ is not correct, yo should integrate a dummy variable say $\displaystyle x'$ from $\displaystyle 0$ to $\displaystyle x$, because the fundamental theorem of calculus is valid for closed intervals $\displaystyle [a,b]$. The notation for improper integrals is a little more messy and confusing, in my opinion

3. Re: Techniques of integration: Trigonometric substitution

I have solved the answer. I used the double-angle trig' identity: $\displaystyle sin^3(x)cos^3(x) \equiv \frac{1}{8}sin^3(2x)$

Oh is it. I used $\displaystyle F(x)$ so "something" equates to "something", instead of starting with "the integral of".

Anyway, why didn't a change-of-variable for $\displaystyle u$ work-out in this problem? It seemed "dooable" when I tried?

Regards

4. Re: Techniques of integration: Trigonometric substitution

Don't doubt yourself! Your original answer is correct. The answer obtained by the second method is correct as well. So what conclusion must you draw? The two functions you obtained,

$\displaystyle F(x) = \frac{1}{48}\cos^3(2x)-\frac{1}{16}\cos(2x)$ and

$\displaystyle G(x) = \frac{1}{6}\cos^6(x)-\frac{1}{4}\cos^4(x)$

must therefore have the same derivative. And indeed they only differ by a constant. According to maple, after simplifying and using trigonometric identities:
$\displaystyle F(x) - G(x) = \frac{1}{24}$

I hope you can see the beauty of math is this example, using integration, you effectively proved that the two functions must differ by some constant. If you are interested you could show directly using trigonometric identities that the two antiderivatives differ by a constant, for full appreciation of this situation.

5. Re: Techniques of integration: Trigonometric substitution

Sword is right
Both functions are correct and differ by a constant.
If you expand the first answer and compare it with the second you will get indeed 2/48 =1/24

6. Re: Techniques of integration: Trigonometric substitution

Originally Posted by SworD
Don't doubt yourself! Your original answer is correct. The answer obtained by the second method is correct as well. So what conclusion must you draw? The two functions you obtained,

$\displaystyle F(x) = \frac{1}{48}\cos^3(2x)-\frac{1}{16}\cos(2x)$ and

$\displaystyle G(x) = \frac{1}{6}\cos^6(x)-\frac{1}{4}\cos^4(x)$

must therefore have the same derivative. And indeed they only differ by a constant. According to maple, after simplifying and using trigonometric identities:
$\displaystyle F(x) - G(x) = \frac{1}{24}$

I hope you can see the beauty of math is this example, using integration, you effectively proved that the two functions must differ by some constant. If you are interested you could show directly using trigonometric identities that the two antiderivatives differ by a constant, for full appreciation of this situation.
Hi,

I think you've just read my mind. Whilst I've been away with coffee and a cigarette in my back-garden, I've just been thinking to myself $\displaystyle \int sin(x)cos(x)\ dx$ has two different but true solutions, but both of which are correct and differ' by a constant!

Such a case induced anxiety because I have my exams coming up soon and if I don't pass - then no University! So this means the examiner can mark either of which I put down in either event.