calculate-0-pi-2-ln-2-sin-d-

Question Number 100967 by mathmax by abdo last updated on 29/Jun/20

calculate \int_{0}^{\frac{π}{2}} \ln (2 + \sin θ) d θ

Answered by mathmax by abdo last updated on 01/Jul/20

I = \int_{0}^{\frac{π}{2}} \ln (2 + \sin θ) d θ \Rightarrow I = \int_{0}^{\frac{π}{2}} (\ln (2) + \ln (1 + \frac{1}{2} \sin θ)) d θ

= \frac{π}{2} \ln (2) + \int_{0}^{\frac{π}{2}} \ln (1 + \frac{1}{2} \sin θ) d θ let f (a) = \int_{0}^{\frac{π}{2}} \ln (1 + asin θ) d θ with 0 < a < 1

we have f^{^{'}} (a) = \int_{0}^{\frac{π}{2}} \frac{\sin θ}{1 + asin θ} = \frac{1}{a} \int_{0}^{\frac{π}{2}} \frac{1 + asin θ - 1}{1 + asin θ} d θ

= \frac{π}{2 a} - \frac{1}{a} \int_{0}^{\frac{π}{2}} \frac{d θ}{1 + asin θ} =_{\tan (\frac{θ}{2}) = t} \frac{π}{2 a} - \frac{1}{a} \int_{0}^{1} \frac{2 dt}{(1 + t^{2}) (1 + a \times \frac{2 t}{1 + t^{2}})}

= \frac{π}{2 a} - \frac{2}{a} \int_{0}^{1} \frac{dt}{1 + t^{2} + 2 at} we have

\int_{0}^{1} \frac{dt}{t^{2} + 2 at + 1} = \int_{0}^{1} \frac{dt}{t^{2} + 2 at + a^{2} + 1 - a^{2}} = \int_{0}^{1} \frac{dt}{{(t + a)}^{2} + 1 - a^{2}}

=_{t + a = \sqrt{1 - a^{2}} u} \int_{\frac{a}{\sqrt{1 - a^{2}}}}^{\frac{1 + a}{\sqrt{1 - a^{2}}}} \frac{\sqrt{1 - a^{2}} du}{(1 - a^{2}) (1 + u^{2})}

= \frac{\sqrt{1 - a} \times \sqrt{1 + a}}{(1 - a) (1 + a)} \int_{\frac{a}{\sqrt{1 - a^{2}}}}^{\frac{1 + a}{\sqrt{1 - a^{2}}}} \frac{du}{1 + u^{2}} = \frac{1}{\sqrt{1 - a^{2}}} {\arctan (\frac{1 + a}{\sqrt{1 - a^{2}}}) - arctn (\frac{a}{\sqrt{1 - a^{2}}})} \Rightarrow

f (a) = \int_{0}^{a} \frac{1}{\sqrt{1 - z^{2}}} \arctan (\frac{1 + z}{\sqrt{1 - z^{2}}}) dz - \int_{0}^{a} \frac{1}{\sqrt{1 - z^{2}}} \arctan (\frac{z}{\sqrt{1 - z^{2}}}) + c

c = f (0) = 0 \Rightarrow

f (a) = \int_{0}^{a} \frac{1}{\sqrt{1 - z^{2}}} \arctan (\frac{1 + z}{\sqrt{1 - z^{2}}}) dz - \int_{0}^{a} \frac{1}{\sqrt{1 - z^{2}}} \arctan (\frac{z}{\sqrt{1 - z^{2}}}) dz

I = \frac{π}{2} \ln (2) + f (\frac{1}{2})

= \frac{π}{2} \ln (2) + \int_{0}^{\frac{1}{2}} \frac{1}{\sqrt{1 - z^{2}}} \arctan (\frac{1 + z}{\sqrt{1 - z^{2}}}) - \int_{0}^{\frac{1}{2}} \frac{1}{\sqrt{1 - z^{2}}} \arctan (\frac{z}{\sqrt{1 - z^{2}}}) dz

rest calculus of this integrals \dots .