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Question Number 32363 by prof Abdo imad last updated on 23/Mar/18

l e t c o n s i d e r t h e f u n c t i o n

f (x, θ) = \int_{x}^{x^{2}} l n (2 + s i n θ c o s t) d t

c a l c u l a t e \frac{\partial f}{\partial x} (x, θ) a n d \frac{\partial f}{\partial θ} (x, θ) .

Commented by prof Abdo imad last updated on 25/Mar/18

\frac{\partial f}{\partial x} (x, θ) = 2 x l n (2 + s i n θ c o s (x^{2})) - l n (2 + s i n θ c o s x)

\frac{\partial f}{\partial θ} (x, θ) = \int_{x}^{x^{2}} \frac{c o s t c o s θ}{2 + c o s t s i n θ} d t c h . t a n (\frac{t}{2}) = u g i v e

\frac{\partial f}{\partial θ} = \int_{t a n (\frac{x}{2})}^{t a n (\frac{x^{2}}{2})} \frac{c o s θ \frac{1 - u^{2}}{1 + u^{2}}}{2 + s i n θ \frac{1 - u^{2}}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}}

= \int_{t a n (\frac{x}{2})}^{t a n (\frac{x^{2}}{2})} \frac{(1 - u^{2}) c o s θ}{(1 + u^{2}) (2 + (1 - u^{2}) s i n θ)} d u

= \int_{t a n (\frac{x}{2})}^{t a n (\frac{x^{2}}{2})} \frac{α - α u^{2}}{(1 + u^{2}) (2 + β - β u^{2})} d u w h i t α = c o s θ

a n d β = s i n θ

= \int_{t a n (\frac{x}{2})}^{t a n (\frac{x^{2}}{2})} \frac{- α u^{2} + α}{(1 + u^{2}) (- β u^{2} + β + 2)} d u l e t d e c o m p o s e

f (u) = \frac{α u^{2} - α}{(1 + u^{2}) (β u^{2} - β - 2)}

= \frac{α u^{2} - α}{β (1 + u^{2}) (u^{2} - {(\sqrt{\frac{β + 2}{β}})}^{2})}

= \frac{a}{(u - \sqrt{\frac{β + 2}{β}})} + \frac{b}{u + \sqrt{\frac{β + 2}{β}}} + \frac{c u + d}{u^{2} + 1} \dots b e

c o n y i n u e d \dots .