let give f(x) =∫_0 ^(π/2) (dt/(1+x tant)) 1) find a simple form of f(x) 2) calculate ∫_0 ^(π/2) ((tant)/((1+xtant)^2 ))dt 3)give the value of ∫_0 ^(π/2) ((tant)/((1+(√3) tant)^2 )) dt .


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Question Number 32040 by abdo imad last updated on 18/Mar/18

$l e t g i v e f (x) = \int_{0}^{\frac{π}{2}} \frac{d t}{1 + x t a n t}$ $1) f i n d a s i m p l e f o r m o f f (x)$ $2) c a l c u l a t e \int_{0}^{\frac{π}{2}} \frac{t a n t}{{(1 + x t a n t)}^{2}} d t$ $3) g i v e t h e v a l u e o f \int_{0}^{\frac{π}{2}} \frac{t a n t}{{(1 + \sqrt{3} t a n t)}^{2}} d t .$
Commented by abdo imad last updated on 21/Mar/18

$c h . t a n t = u g i v e f (x) = \int_{0}^{\infty} \frac{1}{1 + {x u}^{2}} \frac{d u}{1 + u^{2}}$ $= \int_{0}^{\infty} \frac{d u}{(1 + u^{2}) (1 + {x u}^{2})} l e t d e c o m p o s e$ $F (u) = \frac{1}{(1 + u^{2}) (1 + {x u}^{2})} = \frac{a u + b}{1 + u^{2}} + \frac{c u + d}{1 + {x u}^{2}}$ $F (- u) = F (u) \Rightarrow a = c = 0 \Rightarrow F (u) = \frac{b}{1 + u^{2}} + \frac{d}{1 + {x u}^{2}}$ $F (0) = 1 = b + d$ $F (1) = \frac{1}{2 (1 + x)} = \frac{b}{2} + \frac{d}{1 + x} = \frac{b}{2} + \frac{2 d}{2 (1 + x)} \Rightarrow$ $\frac{1}{1 + x} = b + \frac{2 d}{1 + x} \Rightarrow 1 = (1 + x) b + 2 d \Rightarrow$ $1 = (1 + x) b + 2 (1 - b) \Rightarrow 1 = 2 + (x - 1) b \Rightarrow b = \frac{1}{1 - x}$ $d = 1 - b = 1 - \frac{1}{1 - x} = \frac{- x}{1 - x} \Rightarrow$ $F (u) = \frac{1}{(1 - x) (1 + u^{2})} - \frac{x}{(1 - x) (1 + {x u}^{2})}$ $f (x) = \frac{1}{1 - x} \int_{0}^{\infty} \frac{d u}{1 + u^{2}} - \frac{x}{1 - x} \int_{0}^{\infty} \frac{d u}{1 + {x u}^{2}} b u t w e h a v e$ $\int_{0}^{\infty} \frac{d u}{1 + u^{2}} = {[a r c t a n u]}_{0}^{\infty} = \frac{π}{2} a n d c h . \sqrt{x} u = t (w e s u p p o s e x > 0)$ $\int_{0}^{\infty} \frac{d u}{1 + {x u}^{2}} = \int_{0}^{\infty} \frac{1}{1 + t^{2}} \frac{d t}{\sqrt{x}} = \frac{π}{2 \sqrt{x}} \Rightarrow$ $f (x) = \frac{π}{2 (1 - x)} - \frac{π x}{2 \sqrt{x} (1 - x)} = \frac{π}{2 (1 - x)} (1 - \sqrt{x})$ $= \frac{π}{2} \frac{1 - \sqrt{x}}{(1 + \sqrt{x}) (1 - \sqrt{x})} = \frac{π}{2 (1 + \sqrt{x})} .$ $2) w e h a v e f (x) = \int_{0}^{\frac{π}{2}} \frac{d t}{1 + x t a n t} \Rightarrow$ $f^{^{'}} (x) = - \int_{0}^{\frac{π}{2}} \frac{t a n t}{{(1 + x t a n t)}^{2}} d t \Rightarrow$ $\int_{0}^{\frac{π}{2}} \frac{t a n t}{{(1 + x t a n t)}^{2}} d t = - f^{^{'}} (x) b u t$ $f^{^{'}} (x) = - \frac{π}{2} \frac{{(1 + \sqrt{x})}^{^{'}}}{{(1 + \sqrt{x})}^{2}} = - \frac{π}{2} \frac{1}{2 \sqrt{x} {(1 + \sqrt{x})}^{2}} \Rightarrow$ $\int_{0}^{\frac{π}{2}} \frac{t a n t}{{(1 + x t a n t)}^{2}} d t = \frac{π}{4 \sqrt{x} {(1 + \sqrt{x})}^{2}}$ $3) l e t t a k e x = \sqrt{3} w e g e t$ $\int_{0}^{\frac{π}{2}} \frac{t a n t}{{(1 + \sqrt{3} t a n t)}^{2}} d t = \frac{π}{4^{4} \sqrt{3} {(1 +^{4} \sqrt{3})}^{2}} .$
Commented by abdo imad last updated on 21/Mar/18

$x > 0 .$
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