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प्रश्न
Evaluate the definite integral:
`int_0^(pi/2) cos^2 xdx`
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उत्तर
`int_0^(pi/2) cos^2 x dx` ...(i)
= `1/2 int_0^(pi/2) (1 + cos 2x) dx`
`because cos 2x = 1 - 2 cos^2 x`
`=> cos^2 x = 1/2 (1 + cos 2x)`
= `1/2 [x + (sin 2x)/2]_0^(pi/2)`
= `1/2 [pi/2 - (sin pi)/2 - 0 - 1/2 sin 0]`
= `1/2 [pi/2 + 0 - 0]`
= `pi/4`
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