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Question Number 197272 by Erico last updated on 12/Sep/23

How to calculate this integral

{\int_{0}}^{\frac{π}{2}} \frac{\ln (1 + s i n t)}{s i n t} d t

Answered by Mathspace last updated on 12/Sep/23

t a n (\frac{t}{2}) = x \Rightarrow

I = \int_{0}^{1} \frac{l n (1 + \frac{2 x}{1 + x^{2}})}{\frac{2 x}{1 + x^{2}}} \frac{2 d x}{1 + x^{2}}

= \int_{0}^{1} \frac{l n (\frac{1 + x^{2} + 2 x}{1 + x^{2}})}{x} d x

= \int_{0}^{1} \frac{l n {(x + 1)}^{2} - l n (1 + x^{2})}{x} d x

= 2 \int_{0}^{1} \frac{l n (1 + x)}{x} d x - \int_{0}^{1} \frac{l n (1 + x^{2})}{x} d x

\frac{d x}{d x} l n (1 + x) = \frac{1}{1 + x} = \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{n}

\Rightarrow l n (1 + x) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{n + 1} x^{n + 1}

= \sum_{n = 1}^{\infty} {(- 1)}^{n - 1} \frac{x^{n}}{n} \Rightarrow

\int_{0}^{1} \frac{l n (1 + x)}{x} d x = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} \int_{0}^{1} x^{n - 1} d x

= \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n^{2}} = η (2)

= (1 - 2^{1 - 2}) ξ (2) = \frac{1}{2} . \frac{π^{2}}{6} = \frac{π^{2}}{12}

l n (1 + x^{2}) = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} x^{2 n} \Rightarrow

\int_{0}^{1} \frac{l n (1 + x^{2})}{x} d x = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n} \int_{0}^{1} x^{2 n - 1} d x

= \frac{1}{2} \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n - 1}}{n^{2}}

= \frac{1}{2} \frac{π^{2}}{12} = \frac{π^{2}}{24} \Rightarrow

\int_{0}^{\frac{π}{2}} \frac{l n (1 + s i n x)}{s i n x} d x

= 2 . \frac{π^{2}}{12} - \frac{π^{2}}{24} = \frac{3 π^{2}}{24} = \frac{π^{2}}{8}

Answered by universe last updated on 12/Sep/23

I = \int_{0}^{π / 2} \frac{\log (1 + α \sin x) d x}{\sin x}

\frac{d I}{d α} = \int_{0}^{π / 2} \frac{1}{1 + α \sin x} d x

\frac{d I}{d α} = \int_{0}^{π / 2} \frac{\sec^{2} x / 2}{1 + \tan^{2} x / 2 + 2 α \tan x / 2} d x

\frac{d I}{d α} = \int_{0}^{1} \frac{2 d y}{y^{2} + 2 α y + 1} = {\int_{0}}^{1} \frac{2 d y}{{(y + α)}^{2} + (1 - α^{2})}

\frac{d I}{d α} = \frac{2}{\sqrt{1 - α^{2}}} \tan^{- 1} \frac{y + α}{\sqrt{1 - α^{2}}} ∣_{0}^{1}

\frac{d I}{d α} = \frac{2}{\sqrt{1 - α^{2}}} [\tan^{- 1} \frac{1 - α}{\sqrt{1 - α^{2}}} - \tan^{- 1} \frac{α}{\sqrt{1 - α^{2}}}]

\frac{d I}{d α} = \frac{2}{\sqrt{1 - α^{2}}} \tan^{- 1} (\frac{1 / \sqrt{1 - α^{2}}}{1 + (1 + α) α / 1 - α^{2}})

\int \frac{d I}{d α} = \int \frac{2}{\sqrt{1 - α^{2}}} \tan^{- 1} \sqrt{\frac{1 - α}{1 + α}} d α

l e t α = \cos β \Rightarrow d α = - \sin β d β

I = - 2 \int \tan^{- 1} (\sqrt{\frac{1 - \cos β}{1 + \cos β}}) d β

I = - 2 \int \frac{β}{2} d β

I = - \frac{β^{2}}{2} + c

I = - \frac{{(\cos^{- 1} α)}^{2}}{2} + c

l e t α = 0 t h e n I = 0

c = \frac{π^{2}}{8}

I = \frac{π^{2}}{8} - \frac{{(\cos^{- 1} α)}^{2}}{2}

n o w p u t α = 1

I = \frac{π^{2}}{8}

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