Trigonometric Integrals Technique

[whatlifeforme]

Homework Statement

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.

Homework Equations

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

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.

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

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

 

[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
$$-\int_{cos(1)}^{cos(2)} (u^3- u^5)du$$

 

[whatlifeforme]

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

 

[iRaid]

Eventually, you won't 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.

 

[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?

 

[iRaid]

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

 

[SammyS]

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

...

answer 1: $\displaystyle \ \frac{(sinx)^4}{4} - \frac{(sinx)^6}{6} + C_1$
...

answer 2: $\displaystyle \ \frac{-(cosx)^4}{4} + \frac{(cosx)^6}{6} + C_2$

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

(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,

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