# Homework Help: Trigonometric Integrals Technique

1. Mar 14, 2013

### whatlifeforme

1. The problem statement, all variables and given/known data
When integrating a function of the form:

$\displaystyle\int_c^d { (sinx)^{a} * (cosx)^{b}}$

Is this a correct simplification of the rules to evaluate:
1. if exponent on sin or cos is odd, and the other is even, separate out one of the odd's and use
an identity on the even part of what is left of the what was odd function.

2. if exponent on sin and cos are both odd, pick one, pull out one of the sines,
or cosines, and use identity on even part left from original odd function.

3. if exponent on sin and cos are both even , use double angle identity, and
foil.

2. Relevant equations
$\displaystyle\int_c^d { (sinx)^{a} * (cosx)^{b}}$

3. The attempt at a solution
for instance, from rule #2 above.

$\displaystyle\int_1^2 { (sinx)^{3} * (cosx)^{3}}$

Pick either sinx or cosx to manipulate. in this case i choose sinx.

$\displaystyle\int_1^2 { (sinx)^{2} * (sinx) * (cosx)^{3}}$

$\displaystyle\int_1^2 { (1-(cosx)^{2})^{2} * (sinx) * (cosx)^{3}}$

2. Mar 14, 2013

### HallsofIvy

Personally, I would not think of (1) and (2) as separate "rules": If either of a or b is odd you can take out one of that function.

It is, however, important to keep the "dx" with the integral and you have left that out:
$$\int_1^2 sin^3(x)cos^3(x)dx= \int_1^2 sin^2(x) cos^3(x) (sin(x)dx)$$
$$= \int_1^2 (1- cos^2(x))cos^3(x)(sin(x)dx= \int_1^2 (cos^3(x)- cos^5(x))(sin(x)dx)$$
And now the substitution u= cos(x), so that du= -sin(x)dx, gives
$$-\int_{cos(1)}^{cos(2)} (u^3- u^5)du$$

3. Mar 14, 2013

### whatlifeforme

so it does not matter which i choose, sine or cosine, if they both have odd exponents?

4. Mar 14, 2013

### iRaid

Eventually, you wont have to memorize which rule to use and whatever, it'll just come naturally. So yes it doesn't matter if you pick sine or cosine for u.

5. Mar 14, 2013

### whatlifeforme

$\displaystyle\int (sinx)^3 * (cosx)^3$

1. using (cosx)^3

$\displaystyle\int (sinx)^3 * (1-(sinx)^2) * cosx$
$\displaystyle\int (sinx)^3 * cosx - \int (sinx)^5 * cosx$

u=sinx; du=cosx dx

$\int u^3 du - \int u^5 du$

answer 1: $\frac{(sinx)^4}{4} - \frac{(sinx)^6}{6} + c$

2. using (sinx)^3

$\displaystyle\int (sinx)^3 * (cosx)^3 dx$

$\displaystyle\int (sinx)^2 * sinx * (cosx)^3 dx$

$\displaystyle\int (1-(cosx)^2) * sinx * (cosx)^3 dx$

$\displaystyle\int sinx * (cosx)^3 - \int sinx * (cosx)^5 dx$

u=cosx; du=-sinx dx

$-\displaystyle\int u^3 du + \int u^5 du$

answer 2: $\frac{-(cosx)^4}{4} + \frac{(cosx)^6}{6} + c$

i suppose answer 1 and answer 2 are equivalent even though they look different?

6. Mar 14, 2013

### iRaid

I think they end up being the same since $sin^{2}x+cos^{2}x=1$.

7. Mar 14, 2013

### SammyS

Staff Emeritus
(You continue to drop the dx from your integrals even after HallsofIvy has pointed out that error to you.)

( I altered the names of the constants of integration in the above quoted text.)

Answer 1 is equivalent to answer 2 as anti-derivatives of $\displaystyle \ \sin^3(x) \cos^3(x)\ .\$

However, to make the two expressions equal, C1 and C2 must be different. This is because,

$\displaystyle \frac{(sinx)^4}{4} - \frac{(sinx)^6}{6}=\frac{-(cosx)^4}{4} + \frac{(cosx)^6}{6}+\frac{1}{12}$