Anonymous
Anonymous asked in 科學及數學數學 · 9 years ago

Definite Integral

Integrate 1/(1 + sin x) from x = 0 to x = pi/2. (Half angle formula not accepted). Please show steps.

1 Answer

Rating
  • yc
    Lv 5
    9 years ago
    Favorite Answer

    ∫(0->π/2)1/(1 + sin x)dx

    =∫(0->π/2)[(1-sinx)/(cosx)^2]dx

    =∫(0->π/2)[(1-sinx)(secx)^2]dx

    =∫(0->π/2)[(secx)^2-(secx)^2(sinx)]dx

    =∫(0->π/2)[(secx)^2-(secx)(tanx)]dx

    =[tanx - secx](0->π/2)

    =[(sinx-1)/cosx](0->π/2)

    =[(sinx-1)(sinx+1)/(cosx(sinx+1))](0->π/2)

    =[-cosx)/(sinx+1)](0->π/2)

    =0/2-(-1)/1

    =1

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