calculate I = ∫_0 ^(π/2) (x^3 +x)cos^2 xdx and J = ∫_0 ^(π/2) (x^3 +x)sin^2 xdx cslculate I and J .


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Question Number 36415 by abdo.msup.com last updated on 01/Jun/18

$c a l c u l a t e I = \int_{0}^{\frac{π}{2}} (x^{3} + x) {c o s}^{2} x d x a n d$ $J = \int_{0}^{\frac{π}{2}} (x^{3} + x) {s i n}^{2} x d x$ $c s l c u l a t e I a n d J .$
Commented by abdo.msup.com last updated on 04/Jun/18
$we have I +J = ∫_0 ^(π/2) (x^3 +x)dx =[(x^4 /4) +(x^2 /2)]_0 ^(π/2) = (π^4 /(4.2^4 )) +(π^2 /(2.2^2 )) =(π^4 /(64)) +(π^2 /8) also we have I −J = ∫_0 ^(π/2) (x^3 +x)cos(2x)dx =_(2x=t) ∫_0 ^π { ((t/2))^3 +(t/2)}cos(t) (dt/2) =(1/2) ∫_0 ^π { (t^3 /8) +(t/2)}cos(t)dt = (1/(16)) ∫_0 ^π { t^3 +4t)cos(2t)dt by parts ∫_0 ^π { t^3 +4t)cos(2t)=[(1/2)(t^3 +4t)sin(2t)]_0 ^π −(1/2)∫_0 ^π {3t^2 +4}sin(2t)dt =−(1/2){ [ −(1/2)(3t^2 +4)cos(2t)]_0 ^π −∫_0 ^π −(1/2)(6t) cos(2t)dt =(1/4){( 3π^2 +4)−4} −(3/2)∫_0 ^π t cos(2t)dt =(3/4)π^2 −(3/2){ [(1/2)t sin(2t)]_0 ^π −∫_0 ^π (1/2)sin(2t)dt} =((3π^2 )/4) +(3/4) ∫_0 ^π sin(2t)dt =((3π^2 )/4) −(3/8)[cos(2t)]_0 ^π = ((3π^2 )/4) so I +J = (π^4 /(64)) +(π^2 /8) and I −J = ((3π^2 )/4) ⇒ 2I = (π^4 /(64)) + ((7π^2 )/8) ⇒ I = (π^4 /(128)) +((7π^2 )/(16)) and 2J =(π^4 /(64)) +(π^2 /8) −((3π^2 )/4) =(π^4 /(64)) −((5π^2 )/8) ⇒ J = (π^4 /(128)) −((5π^2 )/(16)) .$
$w e h a v e I + J = \int_{0}^{\frac{π}{2}} (x^{3} + x) d x$ $= {[\frac{x^{4}}{4} + \frac{x^{2}}{2}]}_{0}^{\frac{π}{2}} = \frac{π^{4}}{4 . 2^{4}} + \frac{π^{2}}{2 . 2^{2}} = \frac{π^{4}}{64} + \frac{π^{2}}{8}$ $a l s o w e h a v e I - J = \int_{0}^{\frac{π}{2}} (x^{3} + x) c o s (2 x) d x$ $=_{2 x = t} \int_{0}^{π} {{(\frac{t}{2})}^{3} + \frac{t}{2}} c o s (t) \frac{d t}{2}$ $= \frac{1}{2} \int_{0}^{π} {\frac{t^{3}}{8} + \frac{t}{2}} c o s (t) d t$ $= \frac{1}{16} \int_{0}^{π} {t^{3} + 4 t) c o s (2 t) d t b y p a r t s$ $\int_{0}^{π} {t^{3} + 4 t) c o s (2 t) = {[\frac{1}{2} (t^{3} + 4 t) s i n (2 t)]}_{0}^{π}$ $- \frac{1}{2} \int_{0}^{π} {3 t^{2} + 4} s i n (2 t) d t$ $= - \frac{1}{2} {{[- \frac{1}{2} (3 t^{2} + 4) c o s (2 t)]}_{0}^{π}$ $- \int_{0}^{π} - \frac{1}{2} (6 t) c o s (2 t) d t$ $= \frac{1}{4} {(3 π^{2} + 4) - 4} - \frac{3}{2} \int_{0}^{π} t c o s (2 t) d t$ $= \frac{3}{4} π^{2} - \frac{3}{2} {{[\frac{1}{2} t s i n (2 t)]}_{0}^{π} - \int_{0}^{π} \frac{1}{2} s i n (2 t) d t}$ $= \frac{3 π^{2}}{4} + \frac{3}{4} \int_{0}^{π} s i n (2 t) d t$ $= \frac{3 π^{2}}{4} - \frac{3}{8} {[c o s (2 t)]}_{0}^{π} = \frac{3 π^{2}}{4} s o$ $I + J = \frac{π^{4}}{64} + \frac{π^{2}}{8} a n d I - J = \frac{3 π^{2}}{4} \Rightarrow$ $2 I = \frac{π^{4}}{64} + \frac{7 π^{2}}{8} \Rightarrow I = \frac{π^{4}}{128} + \frac{7 π^{2}}{16} a n d$ $2 J = \frac{π^{4}}{64} + \frac{π^{2}}{8} - \frac{3 π^{2}}{4} = \frac{π^{4}}{64} - \frac{5 π^{2}}{8} \Rightarrow$ $J = \frac{π^{4}}{128} - \frac{5 π^{2}}{16} .$
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