int_0^(pi/2)sin^3x=?

1 Answer
Apr 30, 2018

The definite integral is 2/3
0.6666666666666667

Explanation:

int_(0)^(pi/2) sin^3x
Now,
=int(-cos^2x) sin(x)dx
Substitute,
color(grey)(u=cos(x)->dx=-((1)/(sinx))dx
Therefore,
=int(u^2-1)dx
Applying linearity,
=int u^2du-int 1du ---(1)

Now, applying power rule in,
int u^2du
=u^3/3 ---(2)
Now, applying constant rule in,
=int 1du
=u ---(3)
Putting (2) and (3) in (1),
intu^2du-int1du
=u^3/3-u
Undo substitution in,
color(grey)(u=cos(x))
Therefore,
=(cos^3x)/(3)-cosx
therefore =cos^3x/(3)-cosx+C