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x = 2\tan ^{-1}\left( t\right) \Rightarrow t=\tan \left( x / 2\right) \\[br]\dfrac {dx} {dt}=\dfrac {2} {1+t^{2}}\Rightarrow \dfrac {dt} {dx}=\dfrac {1} {2}\left( 1+t^{2}\right) \\[br]\sin \left( x\right) =2\sin \left( x / 2\right) \cos \left( x / 2\right) =2\dfrac {t} {\sqrt {1+t^{2}}}\dfrac {1} {\sqrt {1+t^{2}}}=\dfrac {2t} {1+t^{2}}[br]
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The integrand turns into a rational function with quadratic denominator. After completing the square and substituting the squared term, it becomes a standard inverse tan integral, because 1 – a2 is positive (since 0 < a < 1).
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The result of integration is the difference between two inverse tan functions, A and B. This is simplified by considering A – B as tan-1(tan(A – B)) (notice A – B is within the range of tan-1(x)) and simplifying using the multiple angle formula on the inner bracket.
\displaystyle\int _{0}^{\pi/2}\dfrac {dx} {1+a\sin \left( x\right) } = \displaystyle\int _{\tan \left( 0\right) }^{\tan \left(\pi/4\right) }\dfrac {\dfrac {2dt} {1+t^{2}}} {1+\dfrac {2at} {1+t^{2}}} = \displaystyle\int _{0}^{1}\dfrac {2dt} {t^{2}+2at+1} \\[br]= \displaystyle\int _{0}^{1}\dfrac {2dt} {\left( t+a\right) ^{2}+1-a^{2}} = 2\displaystyle\int _{a}^{1+a}\dfrac {du} {u^{2}+\left( 1-a^{2}\right) } \\[br]=\dfrac {2} {\sqrt {1-a^{2}}}\left( \tan ^{-1}\left(\dfrac {1+a} {\sqrt {1-a^{2}}}\right) - \tan ^{-1}\left( \dfrac {a} {\sqrt {1-a^{2}}}\right) \right) \\[br]= \dfrac {2} {\sqrt {1-a^{2}}} \tan ^{-1}\left( \dfrac {\dfrac {1+a-a} {\sqrt {1-a^{2}}}} {1+\dfrac {a\left( 1+a\right) } {1-a^{2}}}\right) = \dfrac {2} {\sqrt {1-a^{2}}} \tan ^{-1}\left( \dfrac {\sqrt {1-a^{2}}} {1-a^{2}+a+a^{2}}\right) \\[br]= \dfrac {2} {\sqrt {1-a^{2}}} \tan ^{-1}\left( \dfrac {\sqrt {1-a}} {\sqrt {1+a}}\right)
I_{n+1}+2I_{n}=\displaystyle\int _{0}^{\pi/2}\dfrac {\sin ^{n+1}\left( x\right) +2\sin ^{n}\left( x\right) } {2+\sin \left( x\right) }dx = \displaystyle\int _{0}^{\pi/2}\dfrac {\sin ^{n}\left( x\right)\left(\sin(x) + 2\right) } {2+\sin \left( x\right) }dx \\[br]= \displaystyle\int _{0}^{\pi/2}\sin ^{n}\left( x\right) dx \Rightarrow I_{n+1} = \displaystyle\int _{0}^{\dfrac {\pi } {2}}\sin ^{n}\left( x\right)dx - 2I_n
I_3 = \displaystyle\int _{0}^{\pi/2}\sin ^{2}\left( x\right)dx - 2\left( \displaystyle\int _{0}^{\pi/2}\sin\left( x\right)dx - 2\left(\displaystyle\int _{0}^{\pi/2}dx-2I_{0}\right)\right) \\[br]= \dfrac{1}{2}\displaystyle\int _{0}^{\pi/2} \left(1-\cos \left( 2x\right)\right) dx - 2\displaystyle\int _{0}^{\pi/2}\sin\left( x\right)dx +4\displaystyle\int _{0}^{\pi/2}dx-8I_{0} \\[br]= \dfrac {1} {2}\left( \dfrac {\pi } {2}-\dfrac {1} {2}\left[ \sin \left( 2x\right) \right]_0^{\pi / 2}\right) + 2\left[ \cos \left( x\right) \right] _{0}^{\pi / 2} + 2\pi - 8I_0 \\[br]= \dfrac {\pi } {4} - 2 + 2\pi -\dfrac {8} {2}\displaystyle\int _{0}^{\pi / 2}\dfrac {dx} {1+\dfrac {1} {2}\sin \left( x\right) } \\[br]= \dfrac {9\pi } {4} - 2 - 4\left( \dfrac {2} {\sqrt {1-\frac{1}{4}}} \tan ^{-1}\left( \dfrac {\sqrt {1-\frac{1}{2}}} {\sqrt {1+\frac{1}{2}}}\right)\right) = \dfrac {9\pi } {4} -2 - \dfrac {16} {\sqrt {3}}\tan ^{-1}\left(\dfrac {1} {\sqrt {3}}\right) \\[br]= \dfrac {9\pi } {4} - 2 - \dfrac {8\pi} {3\sqrt {3}}= \left(\dfrac {9} {4} - \dfrac {8\sqrt {3}} {9}\right)\pi - 2[br][br]
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