x-1-csc-x-dx-

Question Number 174628 by cortano1 last updated on 06/Aug/22

Ω = \int \frac{x}{1 + \csc x} d x = ?

Commented by infinityaction last updated on 06/Aug/22

\int \frac{x}{1 + \csc x} d x = \int \frac{x s in x}{1 + \sin x} d x

= \int x \frac{1 + \sin x - 1}{1 + \sin x} d x = \int (x - \frac{x}{1 + \sin x}) d x

Ω = \frac{x^{2}}{2} - \int \frac{x}{\underset{I}{\underset{⏟}{1 + \sin x}}}

I = \int \frac{x}{1 + \sin x} \Rightarrow I = \int \frac{x}{1 + \cos (\frac{π}{2} - x)}

I = \int \frac{x}{{2 \cos}^{2} (\frac{π}{4} - \frac{x}{2})}

\frac{π}{4} - \frac{x}{2} = y \Rightarrow \frac{d x}{2} = - d y

I = - \int \frac{(\frac{π}{2} - 2 y) d y}{\cos^{2} y}

I = 2 \int y \sec^{2} y d y - \frac{π}{2} \int \sec^{2} y d y

I = 2 y \tan y - \log ∣ \sin y ∣ - \frac{π}{2} \tan y + C

Ω = \frac{x^{2}}{2} + \log ∣ \sin y ∣ + \frac{π}{2} \tan y - 2 y \tan y + C

h e r e y = \frac{π}{4} - \frac{x}{2}

Commented by Tawa11 last updated on 09/Aug/22

Great sir

Answered by aleks041103 last updated on 06/Aug/22

\int \frac{x}{1 + c s c x} d x = \int \frac{x s i n x}{1 + s i n x} d x =

= \int x \frac{1 + s i n x - 1}{1 + s i n x} d x = \int (x - \frac{x}{1 + s i n x}) d x =

= \frac{x^{2}}{2} - \int \frac{x d x}{1 + s i n x} = \frac{x^{2}}{2} + I

z = t a n (x / 2) \Rightarrow x = 2 a r c t a n (z) \Rightarrow d x = \frac{2 d z}{1 + z^{2}}

s i n (x) = 2 s i n (x / 2) c o s (x / 2) =

= 2 t a n (x / 2) {c o s}^{2} (x / 2) =

= \frac{2 t a n (x / 2)}{1 + {t a n}^{2} (x / 2)} = \frac{2 z}{1 + z^{2}}

\Rightarrow \frac{1}{1 + s i n x} = \frac{1 + z^{2}}{{(1 + z)}^{2}}

\Rightarrow I = 4 \int a r c t a n (z) (- \frac{d z}{{(1 + z)}^{2}}) =

= 4 \int a r c t a n (z) d (\frac{1}{1 + z}) =

= \frac{4 a r c t a n (z)}{1 + z} - 4 \int \frac{d z}{(1 + z) (1 + z^{2})} =

= \frac{2 x}{1 + t a n (x / 2)} + 2 \int \frac{- 2 d z}{(1 + z) (1 + z^{2})}

\frac{- 2}{(1 + z) (1 + z^{2})} = \frac{a}{1 + z} + \frac{b z + c}{1 + z^{2}}

\Rightarrow a + {a z}^{2} + b z + c + {b z}^{2} + c z = - 2

a + c = - 2, b + c = 0, a + b = 0 \Rightarrow - a = - c = b = 1

\Rightarrow \int \frac{- 2 d z}{(1 + z) (1 + z^{2})} = \int (\frac{z - 1}{1 + z^{2}} - \frac{1}{1 + z}) d z =

= \int \frac{z d z}{1 + z^{2}} - \int \frac{d z}{1 + z^{2}} - \int \frac{d z}{1 + z} =

= \frac{1}{2} \int \frac{d (1 + z^{2})}{1 + z^{2}} - a r c t a n (z) - l n (1 + z) =

= \frac{1}{2} l n (1 + z^{2}) - l n (1 + z) - a r c t a n (z) + C =

= \frac{1}{2} l n (\frac{1 + z^{2}}{{(1 + z)}^{2}}) - a r c t a n (z) + C =

= - \frac{1}{2} l n (1 + s i n x) - \frac{1}{2} x + C

\Rightarrow I = \frac{2 x}{1 + t a n (x / 2)} - x - l n (1 + s i n x) + C

\Rightarrow \int \frac{x}{1 + c s c x} d x = \frac{x^{2}}{2} + \frac{2 x}{1 + t a n (x / 2)} - x - l n (1 + s i n x) + C

Commented by Tawa11 last updated on 09/Aug/22

Great sir

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