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Question Number 105393 by john santu last updated on 28/Jul/20

\underset{- π}{\int^{π}} \frac{x \sin x d x}{(1 + x + \sqrt{1 + x^{2}}) \sqrt{3 + \sin^{2} x}}

Answered by maths mind last updated on 29/Jul/20

A = \int_{- π}^{π} \frac{(1 + x - \sqrt{1 + x^{2}}) s i n (x)}{2 \sqrt{3 + {s i n}^{2} (x)}} \dots . . j u s t * \frac{(1 + x - \sqrt{1 + x^{2}})}{(1 + x - \sqrt{1 + x^{2})}}

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

W r i t e i n t o o d d f u n c t i o n / + s o m t h i n g

f (- x) = - f (x), f = \frac{(1 - \sqrt{1 + x^{2})} s i n (x)}{2 \sqrt{3 + {s i n}^{2} (x)}}

\int_{- a}^{a} f (x) d x = 0 \Rightarrow

A = 2 \int_{0}^{π} \frac{x s i n (x)}{\sqrt{3 + {s i n}^{2} (x)}} d x \dots x = π - t \Rightarrow

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

\Rightarrow 4 \int_{0}^{π} \frac{t s i n (t)}{\sqrt{3 + {s i n}^{2} (t)}} d t = \int \frac{2 π s i n (s)}{\sqrt{3 + {s i n}^{2} (s)}} d s e a s y

{s i n}^{2} (t) = 1 - {c o s}^{2} (t), x = c o s (s) \Leftrightarrow 2 π \int_{- 1}^{1} \frac{d x}{\sqrt{4 - x^{2}}}

A = π \int_{- 1}^{1} \frac{d x}{\sqrt{4 - x^{2}}}