solve-I-0-1-Arcsin-x-1-x-x-2-dx-

Question Number 150356 by mnjuly1970 last updated on 11/Aug/21

s o l v e \dots

I := \int_{0}^{1} \frac{Arcsin (\sqrt{x})}{1 - x + x^{2}} d x = ?

Answered by Lordose last updated on 12/Aug/21

Ω = \int_{0}^{1} \frac{\sin^{- 1} (\sqrt{x})}{1 - x + x^{2}} dx \overset{x = u^{2}}{=} 2 \int_{0}^{1} \frac{{usin}^{- 1} (u)}{1 - u^{2} + u^{4}} du

Ω = 2 \int_{0}^{1} \frac{{usin}^{- 1} (u)}{1 - u^{2} + u^{4}} du \overset{u = \sin (y)}{=} \int_{0}^{\frac{\boldsymbol π}{2}} \frac{ysin (2 y)}{\cos^{2} (y) + \sin^{4} (y)} dy

Ω \overset{y = \frac{x}{2}}{=} 2 \int_{0}^{\boldsymbol π} \frac{xsin (x)}{\cos^{2} (x) + 3} dx

Ω = \frac{1}{2} \int_{0}^{\boldsymbol π} \frac{xsin (x)}{\cos^{2} (x) + 3} dx = \frac{1}{2} \int_{0}^{\boldsymbol π} \frac{(\boldsymbol π - x) \sin (x)}{\cos^{2} (x) + 3} dx

Ω = \frac{\boldsymbol π}{2} \int_{0}^{\boldsymbol π} \frac{\sin (x)}{\cos^{2} (x) + 3} dx \overset{u = \cos (x)}{=} \frac{\boldsymbol π}{2} \int_{- 1}^{1} \frac{1}{u^{2} + 3} du

Ω = ∣ \frac{\boldsymbol π}{2 \sqrt{3}} \tan^{- 1} (\frac{u}{\sqrt{3}}) ∣_{- \frac{\boldsymbol π}{4}}^{\frac{\boldsymbol π}{4}} = \frac{\boldsymbol π^{2}}{6 \sqrt{3}}

Answered by mindispower last updated on 11/Aug/21

f (x) = 1 - x + x^{2}

f (1 - x) = 1 - (1 - x) + {(1 - x)}^{2} = 1 - x + x^{2}

x \to 1 - x

I = \int_{0}^{1} \frac{a r c s i n \sqrt{(1 - x)}}{1 - x + x^{2}} d x

a r c s i n (a) + a r c s i n (b) = a r c s i n (a \sqrt{1 - b^{2}} + b \sqrt{1 - a^{2}})

a r c s i n (\sqrt{1 - x}) + a r c s i n (\sqrt{x}) = a r c s i n ((1 - x) \sqrt{1 - \sqrt{x^{2}} + \sqrt{x} . \sqrt{1 - (\sqrt{{1 - x)}^{2}}})}

x \in [0, 1 [\sqrt{{(1 - x)}^{2}} = 1 - x

= a r c s i n (1 - x + x) = a r c s i n (1) = \frac{π}{2}

2 I = \int_{0}^{1} \frac{a r c s i n (\sqrt{1 - x})}{1 - x + x^{2}} d x + \int_{0}^{1} \frac{a r c s i n (\sqrt{x})}{1 - x + x^{2}} d x

2 I = \int_{0}^{1} \frac{a r c s i n (\sqrt{x}) + a r c s i m (\sqrt{1 - x)}}{1 - x + x^{2}} d x

= \frac{π}{2} \int_{0}^{1} \frac{d x}{1 - x + x^{2}} = \frac{π}{2} \int_{0}^{1} \frac{d x}{{(x - \frac{1}{2})}^{2} + \frac{3}{4}}

= \frac{π}{2} [\frac{2}{\sqrt{3}} a r c t a n {(\frac{2}{\sqrt{3}} (x - \frac{1}{2})]}_{0}^{1}

= \frac{π}{\sqrt{3}} . \frac{π}{3} = \frac{2 π^{2}}{3 \sqrt{3}}

I = \frac{π^{2}}{3 \sqrt{3}}

Answered by Kamel last updated on 11/Aug/21

I = 2 \int_{0}^{1} \frac{t A r c s i n (t)}{1 - t^{2} + t^{4}} d t = 2 \int_{0}^{\frac{π}{2}} \frac{u s i n (u) c o s (u) d u}{{c o s}^{2} (u) + {s i n}^{4} (u)}

= \int_{0}^{π} \frac{u s i n (u) d u}{3 + {c o s}^{2} (u)} = \int_{0}^{π} \frac{(π - t) s i n (t)}{3 + {c o s}^{2} (t)} d t

∴ I = \frac{π}{2} \int_{0}^{π} \frac{s i n (u) d t}{3 + {c o s}^{2} (t)} = π \int_{0}^{\frac{π}{2}} \frac{s i n (t) d t}{3 + {c o s}^{2} (t)} \overset{y = c o s (t)}{=} π \int_{0}^{1} \frac{d y}{3 + y^{2}} = \frac{π}{\sqrt{3}} A r c t g (\frac{1}{\sqrt{3}})

= \frac{π^{2}}{6 \sqrt{3}}

Facebook Tweet Pin