Question #65700

Question

1058 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-09T22:04:05

#pi/4#

#int_0^(pi/2)sin^2thetacolor(white).d theta#

Use the trigonometric identity #sin^2theta=1/2(1-cos2theta)#:

#=1/2int_0^(pi/2)(1-cos2theta)d theta#

Integrating both terms separately:

#=1/2(theta-1/2sin2theta)|_0^(pi/2)#

Evaluating:

#=1/2[pi/2-1/2sinpi-(0-1/2sin0)]#

#=pi/4#