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Evaluate-a-1-2-lnx-2-dx-b-0-pi-6-sin-2-x-cos-3-xdx-

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Question Number 66016 by Rio Michael last updated on 07/Aug/19
Evaluate a. ∫_1 ^2 (lnx)^2 dx b. ∫_0 ^(π/6) sin^2 x cos^3 xdx
Evaluatea.∫12(lnx)2dxb.∫0π6sin2xcos3xdx
Commented by Prithwish sen last updated on 07/Aug/19
b.∫_0 ^(π/6) sin^2 x(1−sin^2 x)cosxdx now put sinx=u⇒cosx dx=du ∫_0 ^(1/2) u^2 (1−u^2 )du =∫_0 ^(1/2) (u^2 −u^4 )du =[(1/3)u^3 −(1/5)u^5 ]_0 ^(1/2) =(1/(24)) − (1/(160)) =((17)/(480)) please check.
b.∫0π6sin2x(1−sin2x)cosxdxnowputsinx=u⇒cosxdx=du∫012u2(1−u2)du=∫012(u2−u4)du=[13u3−15u5]012=124−1160=17480pleasecheck.
Commented by Prithwish sen last updated on 07/Aug/19
a.lnx = t⇒(dx/x) =dt⇒dx=e^t dt ∫_0 ^(ln2) t^2 e^t dt= [t^2 e^t −2te^t +2e^t ]_0 ^(ln2) =2(ln2)^2 −4ln2+2 please check.
a.lnx=t⇒dxx=dt⇒dx=etdt∫0ln2t2etdt=[t2et−2tet+2et]0ln2=2(ln2)2−4ln2+2pleasecheck.
Commented by kaivan.ahmadi last updated on 07/Aug/19
∫(1−cos^2 x)cos^3 xdx=∫(cos^3 xdx−cos^5 x)dx= ∫cosx(1−sin^2 x−(1−sin^2 x)^2 )dx= ∫cosx(1−sin^2 x−1+2sin^2 x−sin^4 x)dx= ∫cosx(sin^2 x−sin^4 x)dx set u=sinx⇒du=cosxdx ∫(u^2 −u^4 )du=(u^3 /3)−(u^5 /5)=((sin^3 x)/3)−((sin^5 x)/5)
∫(1−cos2x)cos3xdx=∫(cos3xdx−cos5x)dx=∫cosx(1−sin2x−(1−sin2x)2)dx=∫cosx(1−sin2x−1+2sin2x−sin4x)dx=∫cosx(sin2x−sin4x)dxsetu=sinx⇒du=cosxdx∫(u2−u4)du=u33−u55=sin3x3−sin5x5
Commented by mathmax by abdo last updated on 07/Aug/19
a) changement lnx =t give ∫_1 ^2 (lnx)^2 dx =∫_0 ^(ln(2)) t^2 e^t dt by parts ∫_0 ^(ln(2)) t^2 e^t dt =[t^2 e^t ]_0 ^(ln(2)) −∫_0 ^(ln(2)) 2t e^t dt =2(ln2)^2 −2 { [te^t ]_0 ^(ln(2)) −∫_0 ^(ln(2)) e^t dt} =2(ln(2))^2 −2{2ln(2)−(2−1)} =2(ln(2))^2 −4ln(2) +2
a)changementlnx=tgive∫12(lnx)2dx=∫0ln(2)t2etdtbyparts∫0ln(2)t2etdt=[t2et]0ln(2)−∫0ln(2)2tetdt=2(ln2)2−2{[tet]0ln(2)−∫0ln(2)etdt}=2(ln(2))2−2{2ln(2)−(2−1)}=2(ln(2))2−4ln(2)+2
Commented by mathmax by abdo last updated on 07/Aug/19
b) let I =∫_0 ^(π/6) sin^2 x cos^3 x ⇒I =∫_0 ^(π/6) (sinxcosx)^2 cosxdx =(1/4)∫_0 ^(π/6) sin^2 2x cosx dx =(1/4) ∫_0 ^(π/6) ((1−cos(4x))/2) cosx dx =(1/8) ∫_0 ^(π/6) (cosx−cos(4x)cosx)dx =(1/8) ∫_0 ^(π/6) cosxdx −(1/8) ∫_0 ^(π/6) cos(4x)cosxdx =(1/8).(1/2) −(1/(16)) ∫_0 ^(π/6) (cos(5x)+cos3x)dx =(1/(16))−(1/(16))[(1/5)sin(5x)+(1/3)sin(3x)]_0 ^(π/6) =(1/(16))−(1/(16)){(1/(10)) +(1/3)} =(1/(16))−(1/(160))−(1/(48)) =....
b)letI=∫0π6sin2xcos3x⇒I=∫0π6(sinxcosx)2cosxdx=14∫0π6sin22xcosxdx=14∫0π61−cos(4x)2cosxdx=18∫0π6(cosx−cos(4x)cosx)dx=18∫0π6cosxdx−18∫0π6cos(4x)cosxdx=18.12−116∫0π6(cos(5x)+cos3x)dx=116−116[15sin(5x)+13sin(3x)]0π6=116−116{110+13}=116−1160−148=….
Commented by Rio Michael last updated on 07/Aug/19
Thanks sirs
Thankssirs
Answered by Tanmay chaudhury last updated on 07/Aug/19
b.∫_0 ^(π/6) sin^2 x× (1−sin^2 x)d(sinx) ∣((sin^3 x)/3)−((sin^5 x)/5)∣_0 ^(π/6) =((((1/2))^3 )/3)−((((1/2))^5 )/5) (1/8)((1/3)−(1/(20))) =((17)/(480))
b.∫0π6sin2x×(1−sin2x)d(sinx)∣sin3x3−sin5x5∣0π6=(12)33−(12)5518(13−120)=17480

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