Hey guys! Today we're diving deep into a super interesting trigonometric integral: the integral of sin cubed x times cos cubed x. This bad boy, often written as ∫ sin³(x)cos³(x) dx, might look a little intimidating at first glance, but trust me, with the right approach, it becomes totally manageable. We're going to break it down step-by-step, making sure you guys understand every little bit. So, grab your favorite beverage, get comfy, and let's conquer this integral together!

Understanding the Core Problem: Trigonometric Integrals

Before we jump straight into our specific problem, let's chat briefly about trigonometric integrals in general. These are integrals that involve trigonometric functions like sine, cosine, tangent, and their buddies. The key to tackling them often lies in using trigonometric identities to simplify the integrand – that's the function we're integrating. Think of identities as secret codes that can transform complex trig expressions into simpler, more manageable forms. For instance, the Pythagorean identity sin²(x) + cos²(x) = 1 is a lifesaver in many situations. We'll be leaning heavily on these identities to solve our ∫ sin³(x)cos³(x) dx problem. The reason we need these tricks is that direct integration of powers of sine and cosine can be tricky. We don't have a simple rule for integrating sinⁿ(x) or cosⁿ(x) directly when 'n' is a power greater than 1. That's where our clever substitutions and identity manipulations come into play. The goal is always to transform the integral into a form where we can use standard integration rules, like the power rule or the substitution rule.

Breaking Down sin³(x)cos³(x)

Alright, let's get back to our main event: the integral of sin³(x)cos³(x) dx. The first thing you might notice is that both the sine and cosine functions are raised to the power of three. When you have odd powers in a trigonometric integral like this, it's a huge clue that a substitution might be our best friend. Specifically, we want to save one of the trigonometric functions with an odd power and use a substitution for the rest. Let's rewrite our integrand: sin³(x)cos³(x) = sin²(x)cos²(x) * sin(x)cos(x). This is a crucial first step because it isolates a single sin(x) and a single cos(x). Now, consider this: if we let u = sin(x), then du = cos(x) dx. Alternatively, if we let u = cos(x), then du = -sin(x) dx. Both look promising! However, the arrangement we have, sin²(x)cos²(x) * sin(x)cos(x), makes it slightly more convenient to target a substitution. Let's try to make the expression entirely in terms of cosine if we plan to substitute sine, or entirely in terms of sine if we plan to substitute cosine. Since we have sin²(x) and cos²(x) terms, we can use the fundamental Pythagorean identity, sin²(x) + cos²(x) = 1. This identity allows us to convert sin²(x) into (1 - cos²(x)) or cos²(x) into (1 - sin²(x)). This flexibility is gold!

Strategy 1: Substitution with Sine

Let's roll with the idea of using u-substitution. We'll rewrite sin³(x)cos³(x) as sin²(x) * cos³(x) * sin(x). Our goal is to make everything in terms of cosine so that we can substitute u = cos(x). Using the Pythagorean identity, sin²(x) = 1 - cos²(x). Substituting this in, we get: (1 - cos²(x)) * cos³(x) * sin(x). Now, let u = cos(x). This means du = -sin(x) dx, or sin(x) dx = -du. Let's plug these into our integral:

∫ (1 - u²) * u³ * (-du)

This simplifies to:

-∫ (u³ - u⁵) du

See how this transformed? We went from a tricky trig integral to a simple polynomial integral. This is the power of substitution, guys! Now, we just need to integrate the polynomial. We can use the power rule for integration, which states that ∫ uⁿ du = (uⁿ⁺¹)/(n+1) + C.

Integrating u³ gives us u⁴/4, and integrating u⁵ gives us u⁶/6. So, our integral becomes:

• (u⁴/4 - u⁶/6) + C

Don't forget the negative sign out front!

-u⁴/4 + u⁶/6 + C

Finally, we substitute back u = cos(x) to get our answer in terms of x:

cos⁶(x)/6 - cos⁴(x)/4 + C

And there you have it! One way to solve this integral. Pretty neat, right? This method works beautifully because we could easily isolate a sin(x) term (or a cos(x) term) and use the Pythagorean identity to convert the remaining even powers.

Strategy 2: Substitution with Cosine

Now, let's try the other substitution to see if we get the same result. This is a great way to double-check our work and reinforce the concepts. This time, we'll rewrite sin³(x)cos³(x) as sin³(x)cos²(x) * cos(x). Our goal is to make everything in terms of sine so that we can substitute u = sin(x). Using the Pythagorean identity, cos²(x) = 1 - sin²(x). Substituting this in, we get: sin³(x) * (1 - sin²(x)) * cos(x). Now, let u = sin(x). This means du = cos(x) dx. Let's plug these into our integral:

∫ u³ * (1 - u²) * du

This simplifies to:

∫ (u³ - u⁵) du

Notice that this time we don't have a negative sign in front like in the previous method. This is because our du = cos(x) dx directly matched the remaining term, whereas before, du = -sin(x) dx required us to absorb the negative sign. Now, we integrate this polynomial using the power rule:

∫ u³ du = u⁴/4 ∫ u⁵ du = u⁶/6

So, our integral becomes:

u⁴/4 - u⁶/6 + C

Finally, we substitute back u = sin(x) to get our answer in terms of x:

sin⁴(x)/4 - sin⁶(x)/6 + C

Wait a minute! This looks different from our first answer (cos⁶(x)/6 - cos⁴(x)/4 + C). Don't panic! This is where another handy trig identity comes into play: sin²(x) + cos²(x) = 1. Let's see if we can show these two answers are equivalent.

From our first answer, we have cos⁶(x)/6 - cos⁴(x)/4 + C. Let's use cos²(x) = 1 - sin²(x).

cos⁶(x) = (cos²(x))³ = (1 - sin²(x))³ = 1 - 3sin²(x) + 3sin⁴(x) - sin⁶(x) cos⁴(x) = (cos²(x))² = (1 - sin²(x))² = 1 - 2sin²(x) + sin⁴(x)

Substitute these back:

(1 - 3sin²(x) + 3sin⁴(x) - sin⁶(x))/6 - (1 - 2sin²(x) + sin⁴(x))/4 + C

This looks like it's going to get messy, but trust me, if you work through the algebra, you'll find it simplifies to sin⁴(x)/4 - sin⁶(x)/6 + C. The key takeaway here is that trigonometric results often have multiple equivalent forms due to the identities. Both answers are correct!

Using Double Angle Formulas (Alternative Approach)

Another way to approach integrals involving powers of sine and cosine is by using power-reduction formulas, which are derived from the double-angle identities. For example, we know that sin(2x) = 2sin(x)cos(x). We can rewrite our integrand:

sin³(x)cos³(x) = (sin(x)cos(x))³

From the double-angle formula, sin(x)cos(x) = sin(2x)/2. Substituting this in:

(sin(2x)/2)³ = sin³(2x)/8

So our integral becomes:

(1/8) ∫ sin³(2x) dx

Now we need to integrate sin³(u) where u = 2x. To do this, we use the same technique as before: rewrite sin³(u) as sin²(u)sin(u) and use the identity sin²(u) = 1 - cos²(u).

(1/8) ∫ (1 - cos²(u))sin(u) du

Now, let v = cos(u). Then dv = -sin(u) du, so sin(u) du = -dv.

(1/8) ∫ (1 - v²) (-dv)

(1/8) ∫ (v² - 1) dv

Integrating with respect to v:

(1/8) (v³/3 - v) + C

Now substitute back v = cos(u):

(1/8) (cos³(u)/3 - cos(u)) + C

Finally, substitute back u = 2x:

(1/8) (cos³(2x)/3 - cos(2x)) + C

This gives us a third form of the answer! Again, all these forms are equivalent. This method using double-angle formulas can be quite efficient, especially when dealing with higher powers.

Why All These Different Answers?

It's totally normal to get different-looking answers when solving indefinite integrals, especially trigonometric ones. This is because of the existence of trigonometric identities and the constant of integration, 'C'. Remember that the derivative of a constant is zero. So, if two functions differ only by a constant, their derivatives will be the same. This means that if F(x) is an antiderivative of f(x), then F(x) + C is also an antiderivative for any constant C. Furthermore, trig identities like sin²(x) + cos²(x) = 1, cos(2x) = cos²(x) - sin²(x), etc., allow us to rewrite expressions in many different, yet equivalent, ways. So, if you get an answer that looks different from someone else's, don't sweat it! Just check if you can use identities to transform one into the other. The crucial test is always to differentiate your answer and see if you get back the original integrand.

Conclusion

So, there you have it, guys! We've successfully tackled the integral of sin³(x)cos³(x) dx using a couple of different, yet equally valid, methods. We saw how recognizing odd powers is a strong hint for u-substitution and how the Pythagorean identity is our best friend in these scenarios. We also explored how double-angle formulas can provide an alternative route. Remember, practice is key! The more you work through these types of integrals, the more comfortable you'll become with spotting the right strategies and applying the identities. Keep experimenting, keep questioning, and don't be afraid to try different approaches. Happy integrating!