Question

 cos3x * sin2x dx

Answer 1

[tex]1/2 \int\sin5x-sinx} \, dx =1/2 \int\sin5x} \, dx - 1/2 \int\ {sinx} \, dx =

=-cos5x/10+cosx/2

Answer 2

To solve the integral of cos(3x) * sin(2x) with respect to x, we can use the product-to-sum identities and integration by substitution.

Let's break down the integral step by step:

∫cos(3x) * sin(2x) dx

Using the product-to-sum identity: cos(3x) * sin(2x) = (1/2) * [sin(3x + 2x) - sin(3x - 2x)]

Simplifying: ∫(1/2) * [sin(5x) - sin(x)] dx

Now we can integrate term by term: (1/2) * ∫sin(5x) dx - (1/2) * ∫sin(x) dx

For the integral of sin(5x), we can use substitution. Let u = 5x, then du = 5 dx: (1/2) * ∫sin(5x) dx = (1/2) * (1/5) * ∫sin(u) du = (1/10) * ∫sin(u) du

Now integrating sin(u) with respect to u gives us -cos(u): (1/10) * (-cos(u))

Substituting back u = 5x: = (1/10) * (-cos(5x))

For the integral of sin(x), it is a straightforward integral and equals -cos(x): (1/2) * (-cos(x))

Putting everything together, the final result is:

(1/10) * (-cos(5x)) - (1/2) * cos(x) + C

where C is the constant of integration.