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Question Number 83781 by john santu last updated on 06/Mar/20

∫ _0^(π/2)  (1/(4sin^2 x+5cos^2 x)) dx

0π214sin2x+5cos2xdx

Commented by niroj last updated on 06/Mar/20

  ∫_0 ^(π/2)   ((  1)/(4sin^2 x+5cos^2 x))dx  = ∫_0 ^(π/2)  (( sec^2 x dx)/( 4tan^2 x+5))    put tan x=t       sec^2 xdx=dt   if x=(π/2)then t=∞   if x=0 then t=0    =  ∫_0 ^( ∞)   (1/(4t^2 +5))dt    =   (1/4)∫_0 ^( ∞)  (1/(t^2 +(5/4)))dt    = (1/4)[∫ (1/((t)^2 +(((√5)/2))^2 ))dt]_0 ^∞    = (1/4)[ (1/((√5)/2)) tan^(−1)  (t/((√5)/2))]_0 ^∞   = (1/4)[ (2/(√5))tan^(−1) (∞).(2/(√5))−0]  = (1/4).(2/(√5)). (π/2)=  (π/(4(√5)))//.

0π214sin2x+5cos2xdx=0π2sec2xdx4tan2x+5puttanx=tsec2xdx=dtifx=π2thent=ifx=0thent=0=014t2+5dt=1401t2+54dt=14[1(t)2+(52)2dt]0=14[152tan1t52]0=14[25tan1().250]=14.25.π2=π45//.

Commented by john santu last updated on 06/Mar/20

in short cut   = ∫ _0^(π/2)  (dx/(a^2 sin^2 x+b^2 cos^2 x))  = (π/(2ab)) [ a=(√4) = 2 , b =(√5) ]  = (π/(2.2.(√5))) = (π/(4(√5))) ★

inshortcut=0π2dxa2sin2x+b2cos2x=π2ab[a=4=2,b=5]=π2.2.5=π45

Answered by MJS last updated on 06/Mar/20

∫(dx/(4sin^2  x +5cos^2  x))=       [t=tan x → dx=(dt/(t^2 +1)); sin x =(t/(√(t^2 +1))); cos x =(1/(√(t^2 +1)))]  =∫(dt/(4t^2 +5))=((√5)/(10))arctan ((2(√5)t)/5) =  =((√5)/(10))arctan ((2(√5)tan x)/5) +C  snswer is ((π(√5))/(20))

dx4sin2x+5cos2x=[t=tanxdx=dtt2+1;sinx=tt2+1;cosx=1t2+1]=dt4t2+5=510arctan25t5==510arctan25tanx5+Csnswerisπ520

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