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प्रश्न
Evaluate the definite integrals `int_0^pi (x tan x)/(sec x + tan x)dx`
उत्तर
Let I = `int_0^π (xtanx)/(secx + tanx)dx` ...(1)
I = `int_0^π {((π - x)tan(π - x))/(sec(π - x) + tan(π - x))}dx` ...`(int_0^a f(x)dx = int_0^a f(a - x)dx)`
`\implies` I = `int_0^π {(-(π - x)tanx)/(-(secx + tanx))}dx`
`\implies` I = `int_0^π ((π - x)tanx)/(secx + tanx)dx` ...(2)
Adding (1) and (2), we obtain
2I = `int_0^π (πtanx)/(secx + tanx)dx`
`implies` 2I = `πint_0^π (sinx/cosx)/(1/cosx + sinx/cosx)dx`
`implies` 2I = `πint_0^π (sinx + 1 - 1)/(1 + sinx)dx`
`implies` 2I = `πint_0^π 1.dx - πint_0^π 1/(1 + sinx)dx`
`implies` 2I = `π[x]_0^π - πint_0^π (1 - sinx)/(cos^2x)dx`
`implies` 2I = `π^2 - πint_0^π (sec^2x - tanx secx)dx`
`implies` 2I = `π^2 - π[tanx - secx]_0^π`
`implies` 2I = π[tan π – sec π – tan 0 + sec 0]
`implies` 2I = π2 – π[0 – (–1) – 0 + 1]
`implies` 2I = π2 – 2π
`implies` 2I = π(π – 2)
`implies` I = `π/2(π - 2)`
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