Question Number 149273 by john_santu last updated on 04/Aug/21

∫_0 ^(π/4) ((e^(tan x) sin ^2 x)/(cos ^4 x)) dx =?

Answered by Ar Brandon last updated on 04/Aug/21

I=∫_0 ^(π/4) ((e^(tanx) sin^2 x)/(cos^4 x))dx=∫_0 ^(π/4) e^(tanx) sec^2 x∙tan^2 xdx =[e^(tanx) ∙tan^2 x]_0 ^(π/4) −2∫_0 ^(π/4) e^(tanx) sec^2 x∙tanxdx =e−2[e^(tanx) tanx]_0 ^(π/4) +2∫_0 ^(π/4) e^(tanx) sec^2 xdx =e−2e+2[e^(tanx) ]_0 ^(π/4) =−e+2(e−1)=e−2

Commented byAr Brandon last updated on 04/Aug/21

Integration by-part with v′(x)=e^(tanx) sec^2 x

Commented byAr Brandon last updated on 04/Aug/21

Hihihi. Puissant

Commented bypuissant last updated on 30/Aug/21

hummm

Answered by john_santu last updated on 04/Aug/21

Answered by ArielVyny last updated on 04/Aug/21

∫_0 ^(π/4) (e^(tanx) /(cos^2 x))×tan^2 xdx t=tanx→dt=(1/(cos^2 x))dx ∫_0 ^1 e^t ×t^2 dt→ { ((du=e^t →u=e^t )),((v=t^2 →dv=2t)) :} ∫_0 ^1 t^2 e^t dt=[t^2 e^t ]_0 ^1 −2∫_0 ^1 e^t t=e^1 −2∫_0 ^1 e^t tdt { ((du=e^t →u=e^t )),((v=t→dv=1)) :} ∫_0 ^1 te^t dt=[te^t ]_0 ^1 −∫_0 ^1 e^t dt=[e^t (t−1)]_0 ^1 =1 ∫_0 ^(π/4) ((e^(tanx) sin^2 x)/(cos^4 x))dx=e^1 −2