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प्रश्न
Evaluate: `int_0^(pi/2) cos^3x "d"x`
उत्तर
`int_0^(pi/2) cos^3x "d"x = int_0^(pi/2)((cos3x + 3cosx)/4) "d"x`
= `1/4[int_0^(pi/2) cos x "d"x + 3int_0^(pi/2)cos x "d"x]`
= `1/4[[(sin3x)/3]_0^(pi/2) + 3[sin x]_0^(pi/2)]`
= `1/4[1/3(sin (3pi)/2 - sin 0) + 3(sin pi/2 - sin 0)]`
= `1/4[1/3 (-1 - 0) + 3(1 - 0)]`
= `1/4((-1)/3 + 3)`
= `1/4(8/3)`
= `2/3`
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