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Question Number 205873 by universe last updated on 01/Apr/24

\int_{0}^{π} \frac{1}{π^{2}} \frac{x}{\sqrt{1 + \sin^{3} x}} [(3 π \cos x + 4 \sin x) \sin^{2} x + 4] d x

Answered by Berbere last updated on 02/Apr/24

x \to π - x; l e t Ω = i n t e g r a l

Ω = \int_{0}^{π} \frac{1}{π^{2}} . \frac{3 π x c o s (x) {s i n}^{2} (x)}{\sqrt{1 + {s i n}^{3} (x)}} d x + \frac{1}{π^{2}} \int_{0}^{π} \frac{4 (1 + {s i n}^{3} (x)) x}{\sqrt{1 + {s i n}^{3} (x)}} d x

= \frac{1}{π} \int_{0}^{π} \frac{3 c o s (x) {s i n}^{2} (x)}{\sqrt{1 + {s i n}^{3} (x)}} . x + 4 \int_{0}^{π} \sqrt{1 + {s i n}^{3} (x)} x d x

= \frac{1}{π} . A + 4 B

B; x \to π - x; B = \frac{1}{π^{2}} \int_{0}^{π} \sqrt{1 + {s i n}^{3} (x)} (π - x) d x \Rightarrow 2 B = \frac{1}{π} \int_{0}^{π} \sqrt{1 + {s i n}^{3} (x)}

B = \frac{1}{2 π} \int_{0}^{π} \sqrt{1 + {s i n}^{3} (x)} d x

A; {\begin{cases} u^{'} = \frac{3 c o s (x) {s i n}^{2} (x)}{\sqrt{1 + {s i n}^{3} (x)}}; u = 2 \sqrt{1 + {s i n}^{3} (x)} \\ v = x \Rightarrow v^{'} = 1 \end{cases}

A = {[2 x \sqrt{1 + {s i n}^{3} (x)}]}_{0}^{π} - 2 \int_{0}^{π} \sqrt{1 + {s i n}^{3} (x)} d x

= 2 π - 2 \int_{0}^{π} \sqrt{1 + {s i n}^{3} (x)} d x

Ω = \frac{1}{π} (2 π - 2 \int_{0}^{π} \sqrt{1 + {s i n}^{3} (x)} d x) + 4 (\frac{1}{2 π} \int_{0}^{π} \sqrt{1 + {s i n}^{3} (x)} d x)

= 2

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