`int_(0)^((pi)/(2))cos^(2)x dx`

Explanation

माना`" "I=int_(0)^(pi//2)cos^(2)xdx" …(1)"`
`rArr" "I=int_(0)^(pi//2)cos^(2)((pi)/(2)-x)dx`
`" "(because int_(0)^(a)f(x)dx=int_(0)^(a)f(a-x)dx)`
`" "I=int_(0)^(pi/2)sin^(2)xdx" ...(2)"`
समीकरण (1 ) और (2 ) को जोड़ने पर,
`2I=int_(0)^(pi//2)(sin^(2)x+cos^(2)x)dx`
`=int_(0)^(pi//2)1 dx" "(because sin^(2)x+cos^(2)x=1)`
`" "=[x]_(0)^(pi//2)=(pi)/(2)-0" "rArr" "I=(pi)/(4)`