What is the integral of sin3ycos5ydy?

Answer 1

Best way: Use angle addition.

Sin(Ax)Cos(Bx) = (1/2) [sin[sum x] + sin[dif x]], where sum = A+B and dif = A-B

To show this,

Sin(Ax)Cos(Bx) = (1/2) [sin[(A+B) x] + sin[(A-B) x]]

= (1/2) [(sin[Ax]Cos[Bx]+sin[Bx]cos[Ax]) + (sin[Ax]cos[-Bx]+sin[-Bx]cos[Ax])]

Using the facts that cos[-k] = cos[k] and sin[-k] = -sin[k], we have:

(1/2) [(sin[Ax]Cos[Bx]+sin[Bx]cos[Ax]) + (sin[Ax]cos[-Bx]+sin[-Bx]cos[Ax])]

(1/2) [(sin[Ax]Cos[Bx]+sin[Bx]cos[Ax]) + (sin[Ax]cos[Bx]-sin[Bx]cos[Ax])]

(1/2) 2sin[Ax]Cos[Bx]

sin[Ax]Cos[Bx]

So,

Int[Sin(3y)Cos(5y)dy] = (1/2)Int[Sin(8y)-Sin(2y)dy]

= (-1/16) Cos[8y] +1/4 Cos[2y] + C

You would get the same result if you used integration by parts twice and played around with trig identities.