Evaluate the following Integrals:
Let
Note that we can write sin3x = sin2x × sin x
We also have sin2x + cos2x = 1
⇒ sin3x = (1 – cos2x) sin x
So,
Put cos x = t
⇒ –sin(x)dx = dt (Differentiating both sides)
⇒ sin(x)dx = –dt
When x = 0, t = cos 0 = 1
When x = π, t = cos π = -1
So, the new limits are 1 and -1.
Substituting this in the original integral,
Recall