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प्रश्न
Evaluate the following:
`int_0^(pi/2) x^2 cos 2x "d"x`
योग
उत्तर
I = `int_0^(pi/2) x^2 cos 2x "d"x`
u = x2, v = cos2x
u' = 2x, v1 = `(sin 2x)/2`
u'' = 2, v2 = `- (cos 2x)/4`
v3 = `- (sin 2x)/8`
I = `[x^2 (sin 2x)/2 + 2x (cos2x)/4 - 2 (sin 2x)/8]_0^(pi/2)`
= `[0 - pi/4 - 0]`
= `- pi/4`
`int_0^(pi/2) x^2 cos 2x "d"x = - pi/4`
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Bernoulli’s Formula
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