calculate A =∫_0 ^1 (1+x^2 )(√(1−x^2 ))dx −∫


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Question Number 48064 by maxmathsup by imad last updated on 18/Nov/18

$c a l c u l a t e A = \int_{0}^{1} (1 + x^{2}) \sqrt{1 - x^{2}} d x - \int_{0}^{1} (1 - x^{2}) \sqrt{1 + x^{2}} d x$
Commented by maxmathsup by imad last updated on 20/Nov/18
$we have A =H−K H=∫_0 ^1 (1+x^2 )(√(1−x^2 ))dx =_(x=sint) ∫_0 ^(π/2) (1+sin^2 t)cost costdt =∫_0 ^(π/2) cos^2 t dt +∫_0 ^(π/2) cos^2 t sin^2 tdt =(1/2)∫_0 ^(π/2) (1+cos(2t))dt +(1/4) ∫_0 ^(π/2) sin^2 (2t)dt =(π/4) +(1/4)[sin(2t)]_0 ^(π/2) +(1/8) ∫_0 ^(π/2) (1−cos(4t))dt =(π/4) +0 +(π/(16)) −(1/(32)) ∫_0 ^(π/2) cos(4t)dt=((5π)/(16)) −(1/(4.32))[sin(4t)]_0 ^(π/2) =((5π)/(16)) let calculate K=∫_0 ^1 (1−x^2 )(√(1+x^2 ))dx ⇒ K =_(x=sh(t) ) ∫_0 ^(argsh(1)) (1−sh^2 t)ch(t)ch(t)dt = ∫_0 ^(ln(1+(√2))) (1−sh^2 t)ch^2 t dt =∫_0 ^(ln(1+(√2))) (1−((ch(2t)−1)/2))(((1+ch(2t))/2))dt =(1/4)∫_0 ^(ln(1+(√2))) (3−ch(2t))(1+ch(2t))dt =(1/4) ∫_0 ^(ln(1+(√2))) {3+3ch(2t)−ch(2t)−ch^2 (2t)}dt =(1/4) ∫_0 ^(ln(1+(√2))) {3 +2ch(2t) −((1+ch(4t))/2)}dt =(3/4)ln(1+(√2)) +(1/2) ∫_0 ^(ln(1+(√2))) ch(2t)dt −(1/8)ln(1+(√2))−(1/8) ∫_0 ^(ln(1+(√2))) ch(4t)dt =(5/8)ln(1+(√2)) +(1/4)[sh(2t)]_0 ^(ln(1+(√2))) −(1/(32))[sh(4t)]_0 ^(ln(1+(√2))) =(5/8)ln(1+(√2))+(1/4)[((e^(2t) −e^(−2t) )/2)]_0 ^(ln(1+(√2))) −(1/(32))[((e^(4t) −e^(−4t) )/2)]_0 ^(ln(1+(√2))) K=(5/8)ln(1+(√2)) +(1/8)( (1+(√2))^2 −(1/((1+(√2))^2 )))−(1/(64))( (1+(√2))^4 −(1/((1+(√2))^4 ))) A=((5π)/(16)) −K so the value of A is known.$
$w e h a v e A = H - K$ $H = \int_{0}^{1} (1 + x^{2}) \sqrt{1 - x^{2}} d x =_{x = s i n t} \int_{0}^{\frac{π}{2}} (1 + {s i n}^{2} t) c o s t c o s t d t$ $= \int_{0}^{\frac{π}{2}} {c o s}^{2} t d t + \int_{0}^{\frac{π}{2}} {c o s}^{2} t {s i n}^{2} t d t$ $= \frac{1}{2} \int_{0}^{\frac{π}{2}} (1 + c o s (2 t)) d t + \frac{1}{4} \int_{0}^{\frac{π}{2}} {s i n}^{2} (2 t) d t$ $= \frac{π}{4} + \frac{1}{4} {[s i n (2 t)]}_{0}^{\frac{π}{2}} + \frac{1}{8} \int_{0}^{\frac{π}{2}} (1 - c o s (4 t)) d t$ $= \frac{π}{4} + 0 + \frac{π}{16} - \frac{1}{32} \int_{0}^{\frac{π}{2}} c o s (4 t) d t = \frac{5 π}{16} - \frac{1}{4.32} {[s i n (4 t)]}_{0}^{\frac{π}{2}}$ $= \frac{5 π}{16} l e t c a l c u l a t e K = \int_{0}^{1} (1 - x^{2}) \sqrt{1 + x^{2}} d x \Rightarrow$ $K =_{x = s h (t)} \int_{0}^{a r g s h (1)} (1 - {s h}^{2} t) c h (t) c h (t) d t$ $= \int_{0}^{l n (1 + \sqrt{2})} (1 - {s h}^{2} t) {c h}^{2} t d t = \int_{0}^{l n (1 + \sqrt{2})} (1 - \frac{c h (2 t) - 1}{2}) (\frac{1 + c h (2 t)}{2}) d t$ $= \frac{1}{4} \int_{0}^{l n (1 + \sqrt{2})} (3 - c h (2 t)) (1 + c h (2 t)) d t$ $= \frac{1}{4} \int_{0}^{l n (1 + \sqrt{2})} {3 + 3 c h (2 t) - c h (2 t) - {c h}^{2} (2 t)} d t$ $= \frac{1}{4} \int_{0}^{l n (1 + \sqrt{2})} {3 + 2 c h (2 t) - \frac{1 + c h (4 t)}{2}} d t$ $= \frac{3}{4} l n (1 + \sqrt{2}) + \frac{1}{2} \int_{0}^{l n (1 + \sqrt{2})} c h (2 t) d t - \frac{1}{8} l n (1 + \sqrt{2}) - \frac{1}{8} \int_{0}^{l n (1 + \sqrt{2})} c h (4 t) d t$ $= \frac{5}{8} l n (1 + \sqrt{2}) + \frac{1}{4} {[s h (2 t)]}_{0}^{l n (1 + \sqrt{2})} - \frac{1}{32} {[s h (4 t)]}_{0}^{l n (1 + \sqrt{2})}$ $= \frac{5}{8} l n (1 + \sqrt{2}) + \frac{1}{4} {[\frac{e^{2 t} - e^{- 2 t}}{2}]}_{0}^{l n (1 + \sqrt{2})} - \frac{1}{32} {[\frac{e^{4 t} - e^{- 4 t}}{2}]}_{0}^{l n (1 + \sqrt{2})}$ $K = \frac{5}{8} l n (1 + \sqrt{2}) + \frac{1}{8} ({(1 + \sqrt{2})}^{2} - \frac{1}{{(1 + \sqrt{2})}^{2}}) - \frac{1}{64} ({(1 + \sqrt{2})}^{4} - \frac{1}{{(1 + \sqrt{2})}^{4}})$ $A = \frac{5 π}{16} - K s o t h e v a l u e o f A i s k n o w n .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 19/Nov/18

	$= \int_{0}^{1} \sqrt{1 - x^{2}} + x^{2} \sqrt{1 - x^{2}} - \sqrt{1 + x^{2}} + x^{2} \sqrt{1 + x^{2}} d x$ $I_{1} = \int_{0}^{1} \sqrt{1 - x^{2}} d x$ $I_{2} = \int_{0}^{1} \sqrt{1 + x^{2}} d x$ $I_{3} = \int_{0}^{1} x^{2} \sqrt{1 - x^{2}} d x$ $I_{4} = \int_{0}^{1} x^{2} \sqrt{1 + x^{2}} d x$ $I = I_{1} - I_{2} + I_{3} + I_{4}$ $u s e f o r m u l a s a t t a c h i n g h e r e w i t h t l s o l v e . . .$
	Commented by tanmay.chaudhury50@gmail.com last updated on 19/Nov/18

	Commented by tanmay.chaudhury50@gmail.com last updated on 19/Nov/18

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