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प्रश्न
Evaluate : `∫_0^(pi/2) (sinx.cosx)/(1 + sin^4x)`.dx
उत्तर
Let I = `∫_0^(pi/2) (sinx.cosx)/(1 + sin^4x)`.dx
Put `sin^2x = t`
2sinx cosx dx = dt
`sinx cosx dx = dt/2`
When x = 0 , t = 0
`x = pi/2` , t = 1
∴ I = `∫_0^1 (dt/2)/(1 + t^2)`
= `∫_0^1 1/2[dt/(1 + t^2)]`
= `[1/2 (tan^-1 t)]_0^1`
= `1/2[tan^-1 1 - tan^-1 0]`
= `1/2[pi/4 - 0]`
`"I" = pi/8`
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