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# Definite Integral

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

### 1 Answer

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- ycLv 59 years agoFavorite 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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