integrate (pi/6 & 0) (1-cos3t)(sin3t)dt
My calculation:
u=1-cos3t
du/dt=sin3t
du=sin3t dt
t=0, u=1-cos3(0)=1
t=pi/6, u=1-cos3(pi/6)=0.477
integral (1&0.477) u du
(u^2/2)(1&0.477)
then stucked
by "inspection" I get
(1/6)(1 - cos 3t)^2 | from 0 to π/6
= (1/6)(1 - 0)^2 - (1/6)(1 - 1)^2
= 1/6 - 0
= 1/6
I was assuming that your limit went from 0 to π/6
Still not very understand.
Where u get the 1/6?
To integrate the function ∫(π/6 to 0) (1 - cos^3(t))(sin(3t)) dt, follow these steps:
1. Expand the cosine cubed term:
∫(π/6 to 0) (1 - (cos(t))^2)(1 - cos(t))(sin(3t)) dt
2. Apply the identity sin^2(t) = 1 - cos^2(t):
∫(π/6 to 0) (sin^2(t))(1 - cos(t))(sin(3t)) dt
3. Expand further:
∫(π/6 to 0) (sin^2(t) - sin^2(t)cos(t))(sin(3t)) dt
4. Use the product-to-sum formula sin(a)sin(b) = 1/2[cos(a-b) - cos(a+b)] to rewrite the second term:
∫(π/6 to 0) (sin^2(t) - (1/2)[cos(4t) - cos(2t)](sin(3t)) dt
5. Distribute to simplify:
∫(π/6 to 0) (sin^2(t) - (1/2)[(cos(4t))(sin(3t)) - (cos(2t))(sin(3t))]) dt
6. Expand further:
∫(π/6 to 0) (sin^2(t) - (1/2)(cos(4t))(sin(3t)) + (1/2)(cos(2t))(sin(3t))) dt
7. Integrate each term individually:
∫(π/6 to 0) (sin^2(t)) dt - (1/2)∫(π/6 to 0)(cos(4t))(sin(3t)) dt + (1/2)∫(π/6 to 0)(cos(2t))(sin(3t)) dt
8. For the first term, apply the identity sin^2(t) = (1/2)(1 - cos(2t)):
(∫(π/6 to 0)(1/2)(1 - cos(2t)) dt) - (1/2)∫(π/6 to 0)(cos(4t))(sin(3t)) dt + (1/2)∫(π/6 to 0)(cos(2t))(sin(3t)) dt
9. Simplify the first term:
(1/2) ∫(π/6 to 0)(1 - cos(2t)) dt - (1/2)∫(π/6 to 0)(cos(4t))(sin(3t)) dt + (1/2)∫(π/6 to 0)(cos(2t))(sin(3t)) dt
10. Integrate each term separately using basic integration rules.
Please note that the above calculation may require further simplification, but these steps outline the process for integrating the given function.