calculate ∫_0 ^1 (dt/((1+t^2 )^3 ))


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Question Number 66328 by mathmax by abdo last updated on 12/Aug/19

$c a l c u l a t e \int_{0}^{1} \frac{d t}{{(1 + t^{2})}^{3}}$
Commented by mathmax by abdo last updated on 15/Aug/19
$let I =∫_0 ^1 (dt/((1+t^2 )^3 )) changement t=tanθ give I =∫_0 ^(π/4) ((1+tan^2 θ)/((1+tan^2 θ)^3 ))dθ =∫_0 ^(π/4) (dθ/((1+tan^2 θ)^2 )) =∫_0 ^(π/4) cos^4 θdθ =∫_0 ^(π/4) (((1+cos(2θ))/2))^2 dθ =(1/4)∫_0 ^(π/4) (1+2cos(2θ) +((1+cos(4θ))/2))dθ =(1/8)∫_0 ^(π/4) {3 +4cos(2θ)+cos(4θ)}dθ =(3/8)(π/4) +(1/2) ∫_0 ^(π/4) cos(2θ)dθ +(1/8)∫_0 ^(π/4) cos(4θ)dθ =((3π)/(32)) +(1/4)[sin(2θ)]_0 ^(π/4) +(1/(32))[sin(4θ)]_0 ^(π/4) =((3π)/(32)) +(1/4) +0 ⇒ I =((3π)/(32)) +(1/4)$
$l e t I = \int_{0}^{1} \frac{d t}{{(1 + t^{2})}^{3}} c h a n g e m e n t t = t a n θ g i v e$ $I = \int_{0}^{\frac{π}{4}} \frac{1 + {t a n}^{2} θ}{{(1 + {t a n}^{2} θ)}^{3}} d θ = \int_{0}^{\frac{π}{4}} \frac{d θ}{{(1 + {t a n}^{2} θ)}^{2}} = \int_{0}^{\frac{π}{4}} {c o s}^{4} θ d θ$ $= \int_{0}^{\frac{π}{4}} {(\frac{1 + c o s (2 θ)}{2})}^{2} d θ = \frac{1}{4} \int_{0}^{\frac{π}{4}} (1 + 2 c o s (2 θ) + \frac{1 + c o s (4 θ)}{2}) d θ$ $= \frac{1}{8} \int_{0}^{\frac{π}{4}} {3 + 4 c o s (2 θ) + c o s (4 θ)} d θ$ $= \frac{3}{8} \frac{π}{4} + \frac{1}{2} \int_{0}^{\frac{π}{4}} c o s (2 θ) d θ + \frac{1}{8} \int_{0}^{\frac{π}{4}} c o s (4 θ) d θ$ $= \frac{3 π}{32} + \frac{1}{4} {[s i n (2 θ)]}_{0}^{\frac{π}{4}} + \frac{1}{32} {[s i n (4 θ)]}_{0}^{\frac{π}{4}}$ $= \frac{3 π}{32} + \frac{1}{4} + 0 \Rightarrow I = \frac{3 π}{32} + \frac{1}{4}$
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