0-pi-x-2-cos-2-x-xsin-x-cos-x-1-1-xsin-x-2-dx-

Question Number 205248 by universe last updated on 13/Mar/24

\int_{0}^{π} \frac{x^{2} \cos^{2} (x) - x \sin (x) - \cos (x) - 1}{{(1 + x \sin (x))}^{2}} d x

Answered by Berbere last updated on 13/Mar/24

{(\frac{f (x)}{1 + x s i n (x)} + g (x))}^{^{'}} = \frac{x^{2} {c o s}^{2} (x) - x s i n (x) - c o s (x) - 1}{{(1 + x s i n (x))}^{2}}

\frac{f^{'} (1 + x s i n (x)) - (s i n (x) + x c o s (x)) f}{{(1 + x s i n (x))}^{2}} + g^{'} (x) = \frac{x^{2} {c o s}^{2} (x) - x s i n (x) - c o s (x) - 1}{{(1 + x s i n (x))}^{2}}

f = - x c o s (x)

\Rightarrow (- c o s (x) + x s i n (x)) (1 + x s i n (x)) - (s i n (x) + x c o s (x)) (- x c o s (x)

x^{2} {c o s}^{2} (x) + x^{2} {s i n}^{2} (x) + x s i n (x) - c o s (x) = f^{'} (1 + x s i n (x)) - (s i n (x) + x c o s (x)) f

\Rightarrow \frac{f^{'} (1 + x s i n (x)) - (s i n (x) + x c o s (x)) f}{{(1 + x s i n (x))}^{2}} + g^{'}

= \frac{x^{2} {c o s}^{2} (x) - c o s (x) - x s i n (x) - 1 + 2 x s i n (x) + x^{2} {s i n}^{2} (x) + 1}{{(1 + x s i n (x))}^{2}}

= \frac{d}{d x} (\frac{- x c o s (x)}{(1 + x s i n (x))}) + \frac{{(1 + x s i n (x))}^{2}}{{(1 + x s i n (x))}^{2}}

Missing \left or extra \right

= {[- \frac{x c o s (x)}{1 + x s i n (x)} - x]}_{0}^{π} = π - π = 0

Commented by universe last updated on 14/Mar/24

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