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Question Number 104312 by Ar Brandon last updated on 20/Jul/20

\int_{0}^{\frac{π}{4}} \frac{\sqrt{\sin^{2} θ + 2}}{\sin θ} d θ

Answered by mathmax by abdo last updated on 21/Jul/20

I = \int_{0}^{\frac{π}{4}} \frac{\sqrt{\sin^{2} θ + 2}}{\sin θ} d θ \Rightarrow I = \int_{0}^{\frac{π}{4}} \frac{\sin θ \sqrt{1 + \frac{2}{\sin^{2} θ}}}{\sin θ} d θ = \int_{0}^{\frac{π}{4}} \sqrt{1 + \frac{2}{1 - \cos^{2} θ}} d θ

we have 1 + \tan^{2} θ = \frac{1}{\cos^{2} θ} \Rightarrow \cos^{2} θ = \frac{1}{1 + \tan^{2} θ} \Rightarrow 1 - \cos^{2} θ = 1 - \frac{1}{1 + \tan^{2} θ}

= \frac{\tan^{2} θ}{1 + \tan^{2} θ} \Rightarrow \frac{2}{1 - \cos^{2} θ} = 2 \times \frac{1 + \tan^{2} θ}{\tan^{2} θ} \Rightarrow

I = \int_{0}^{\frac{π}{4}} \sqrt{1 + \frac{2 + {2 \tan}^{2} θ}{\tan^{2} θ}} d θ = \int_{0}^{\frac{π}{4}} \sqrt{\frac{{3 \tan}^{2} θ + 2}{\tan^{2} θ}} d θ =_{\tan θ = x}

= \int_{0}^{1} \sqrt{\frac{{3 x}^{2} + 2}{x^{2}}} \frac{dx}{1 + x^{2}} = \int_{0}^{1} \frac{\sqrt{{3 x}^{2} + 2}}{x (1 + x^{2})} dx = \sqrt{3} \int_{0}^{1} \frac{\sqrt{x^{2} + \frac{2}{3}}}{x (1 + x^{2})} dx

cha 7 gement x = \sqrt{\frac{2}{3}} shu give u = argsh (\sqrt{\frac{3}{2}} x)

I = \sqrt{3} \int_{0}^{arhsh (\sqrt{\frac{3}{2}})} \frac{2}{3} \times \frac{chu}{\sqrt{\frac{2}{3}} sh (u) (1 + \frac{2}{3} {sh}^{2} u)} \times \sqrt{\frac{2}{3}} ch (u) du

= 2 \sqrt{3} \int_{0}^{\ln (\sqrt{\frac{3}{2}} + \sqrt{1 + \frac{9}{4}})} \frac{{ch}^{2} u}{shu (3 + {2 sh}^{2} u)} du and this integral can be solved

\dots be continued \dots

Commented by Ar Brandon last updated on 21/Jul/20

wow wow wow ! superb. Thanks so much ��

Commented by Ar Brandon last updated on 21/Jul/20

\frac{{ch}^{2} u}{shu (3 + {2 sh}^{2} u)} = \frac{1 + {sh}^{2} u}{shu (3 + {2 sh}^{2} u)}

\frac{1 + t^{2}}{t (3 + {2 t}^{2})} = \frac{at + b}{{2 t}^{2} + 3} + \frac{c}{t} = \frac{(at + b) t + c ({2 t}^{2} + 3)}{t ({2 t}^{2} + 3)}

t \to 0, c = \frac{1}{3}, a + 2 c = 1, a = \frac{1}{3}, b = 0

\Rightarrow \frac{1 + {sh}^{2} u}{shu (3 + {2 sh}^{2} u)} = \frac{\frac{1}{3} shu}{3 + {2 sh}^{2} u} + \frac{\frac{1}{3}}{shu}

\Rightarrow \int \frac{{ch}^{2} u}{shu (3 + {2 sh}^{2} u)} du = \frac{1}{3} {\int \frac{shu du}{3 + {2 ch}^{2} - 2} + \int \frac{du}{shu}}

= \frac{1}{3} {\frac{1}{\sqrt{2}} \int \frac{d (\sqrt{2} chu)}{1 + {2 ch}^{2} u} + \int \frac{shu du}{{ch}^{2} u - 1}}

= \frac{1}{3} {\frac{1}{\sqrt{2}} Arctan (\sqrt{2} chu) - Arctanh (chu)}