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प्रश्न
Evaluate: `int_0^π x/(1 + sinx)dx`.
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उत्तर
`int_0^π x/(1 + sinx)dx`
Let I = `int_0^π x/(1 + sinx)dx` ...(i)
On using property
`int_0^a f(x)dx = int_0^a f(a - x)dx`
∴ I = `int_0^π (π - x)/(1 + sin(π - x))dx`
I = `int_0^π (π - x)/(1 + sinx)dx` ...(ii)
Adding equations (i) and (ii), we get
2I = `int_0^π π/(1 + sinx)dx`
= `πint_0^π 1/(1 + sinx) xx (1 - sinx)/(1 - sinx)dx` ...[Multiplying and dividing by (1 – sin x)]
= `πint_0^π (1 - sinx)/(1 - sin^2x)dx = πint_0^π (1 - sinx)/(cos^2x)dx`
= `π[int_0^π 1/(cos^2x)dx - int_0^π sinx/(cos^2x)dx]`
= `π[int_0^π sec^2x dx - int_0^π secx tanx dx]`
= `π[[tanx]_0^π - [secx]_0^π]`
= π[0 – (– 1 – 1)]
= 2π
∴ I = `(2π)/2` = π.
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