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0-pi-4-tanh-2x-dx-




Question Number 87861 by Rio Michael last updated on 06/Apr/20
∫_0 ^(π/4)  tanh 2x dx
π40tanh2xdx
Commented by MJS last updated on 06/Apr/20
tanh 2x =((e^(4x) −1)/(e^(4x) +1))  ∫tanh 2x dx=       [t=e^(4x)  → dx=(dt/(4e^(4x) ))]  =(1/4)∫((t−1)/(t(t+1)))dt=(1/2)∫(dt/(t+1))−(1/4)∫(dt/t)=  =(1/2)ln (t+1) −(1/4)ln t =−x+(1/2)ln (e^(4x) +1) +C
tanh2x=e4x1e4x+1tanh2xdx=[t=e4xdx=dt4e4x]=14t1t(t+1)dt=12dtt+114dtt==12ln(t+1)14lnt=x+12ln(e4x+1)+C
Commented by Rio Michael last updated on 06/Apr/20
 sir can we use ′′ integration by parts′′?
sircanweuseintegrationbyparts?
Commented by MJS last updated on 07/Apr/20
I don′t think that would be easier
Idontthinkthatwouldbeeasier
Commented by Rio Michael last updated on 07/Apr/20
i will try it out sir
iwilltryitoutsir

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