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प्रश्न
Evaluate: `int_0^(pi/2) x sin x.dx`
उत्तर
`int_0^(pi/2) x sin x.dx`
= `[x int sinx.dx]_0^(pi/2) - int_0^(pi/2)[d/dx(x) int sin x.dx].dx`
= `[x (- cos x)]_0^(pi/2) - int_0^(pi/2) 1.(- cos x).dx`
= `-[x cosx]_0^(pi/2) + int_0^(pi/2) cosx.dx`
= `-[pi/2 cos pi/2 - 0] + [sinx]_0^(pi/2)`
= `0 + (sin pi/2 - sin 0)`
= 1
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