Question Number 149608 by puissant last updated on 06/Aug/21

.....K=∫(1/(1+sin^2 (x)))dx......

Answered by ArielVyny last updated on 06/Aug/21

sin^2 x=((1−cos(2x))/2) K=∫(1/(1+((1−cos(2x))/2)))=∫(1/((2+1−cos(2x))/2))dx K=∫(2/(3−cos(2x)))dx=2∫(1/(3−cos(2x)))dx=_(t=2x) K=∫(1/(3−cos(t)))dt γ=tan((t/2))→dt=(1/2)(1+γ^2 )dx cost=((1−γ^2 )/(1+γ^2 )) K=2∫(1/(3−((1−γ^2 )/(1+γ^2 ))))×(1/(1+γ^2 ))dγ K=2∫(1/(3(1+γ^2 )−(1−γ^2 )))dγ=2∫(1/(3+3γ^2 −1+γ^2 ))dγ K=2∫(1/(2+4γ^2 ))dγ=∫(1/(1+2γ^2 ))dγ=∫(1/(1+((√2)γ)^2 ))dγ u=(√2)γ→du=(√2)dγ→dγ=((√2)/2)du ((√2)/2)∫(1/(1+u^2 ))du=((√2)/2)arctg(u)+cte=((√2)/2)arctg((√2)γ)+cte K=((√2)/2)arctg((√2)tan((t/2)))=((√2)/2)arctg((√2)tan(x)) K=∫(1/(1+sin^2 x))dx=((√2)/2)arctg((√2)tanx)

Commented bypuissant last updated on 06/Aug/21

propre merci..

Answered by Olaf_Thorendsen last updated on 06/Aug/21

K = ∫(dx/(1+sin^2 (x))) K = ∫((cos^2 x+sin^2 x)/(1+sin^2 (x))) dx K = ∫((1+tan^2 x)/((1/(cos^2 x))+tan^2 (x))) dx K = ∫((1+tan^2 x)/(1+2tan^2 (x))) dx K = (1/( (√2)))∫((d((√2)tanx))/(1+((√2)tan(x))^2 )) K = (1/( (√2)))arctan((√2)tanx)+C

Commented bypuissant last updated on 06/Aug/21

jolie merci..

Answered by nimnim last updated on 16/Aug/21

t=tanx⇒ sinx=(t/( (√(1+t^2 )))) dt=sec^2 x.dx⇒(dt/(1+tan^2 x))=dx ∴ K=∫(1/(1+(t^2 /(1+t^2 ))))×(dt/(1+t^2 ))=∫(1/(1+2t^2 ))dt =(1/( (√2)))tan^(−1) ((√2)t)=(1/( (√2)))tan^(−1) ((√2) tanx)+C

Commented bypuissant last updated on 06/Aug/21

jolie..

Answered by Ar Brandon last updated on 06/Aug/21

K=∫(dx/(1+sin^2 x))=∫((cosec^2 x)/(cosec^2 x+1))dx=∫((cosec^2 x)/(2+cot^2 x))dx =−∫((d(cotx))/(2+cot^2 x))=−(1/( (√2)))arctan(((cotx)/( (√2))))+C