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प्रश्न
Evaluate:
`int_0^(pi/4) sqrt(1 + sin 2x)*dx`
उत्तर
`int_0^(pi/4) sqrt(1 + sin 2x)*dx`
= `int_0^(pi/4) sqrt(sin^2x + cos^2x + 2 sin x cos x)*dx`
= `int_0^(pi/4) sqrt((sinx + cosx)^2)*dx`
= `int_0^(pi/4) (sinx + cosx)*dx`
= `int_0^(pi/4) sinx*dx + int_0^(pi/4) cosx*dx`
= `[ - cos x]_0^(pi/4) + [sin x]_0^(pi/4)`
= `[- cos pi/4 - (- cos 0)] + [sin pi/4 - sin 0]`
= `-(1)/sqrt(2) + 1 + (1)/sqrt(2) - 0`
= 1.
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