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Question Number 54936 by maxmathsup by imad last updated on 14/Feb/19
letf(θ)=∫01x2+2(cosθ)x+1dxwithθ∈R.1)calculatef(θ)2)findthevalueofg(θ)=∫01xsinθx2+2cosθx+1dx
Commented by Abdo msup. last updated on 16/Feb/19
1)wehavef(θ)=∫01x2+2cosθx+1dx=∫01(x+cosθ)2+sin2θdθchang.x+cosθ=sinθtgivef(θ)=∫coatanθcotan(θ2)∣sinθ∣sinθ1+t2dtf(θ)=∣sinθ∣sinθ∫cotan(θ)cotan(θ2)1+t2dtletfind∫1+t2dtchangementt=sh(u)give∫1+t2dt=∫ch(u)ch(u)du=∫ch2(u)du=∫1+ch(2u)2du=u2+14sh(2u)=u2+12ch(u)sh(u)=12argsh(t)+12t1+t2+c=12ln(t+1+t2)+t21+t2⇒∫cotan(θ)cotan(θ2)1+t2dt=12[ln(t+1+t2)+t1+t2]cotanθcotan(θ2)=12{ln(cotan(θ2)+1+cotan2(θ2))+cotan(θ2)1+cotan2(θ2)−ln(cotanθ+1+cotanθ−cotanθ1+cotan2θ)
Answered by tanmay.chaudhury50@gmail.com last updated on 15/Feb/19
∫x2+2xcosθ+1dx∫(x+cosθ)2+sin2θdx(x+cosθ)2(x+cosθ)2+sin2θ+sin2θ2(x+cosθ)2+sin2θ+csoansweris∣(x+cosθ)2(x+cosθ)2+sin2θ+sin2θ2(x+cosθ)2+sin2θ∣01[{(1+cosθ)2(1+cosθ)2+sin2θ+sin2θ2(1+cosθ)2+sin2θ}−{cosθ2cos2θ+sin2θ+sin2θ2×cos2θ+sin2θ}][(1+cosθ)21+2cosθ+1+sin2θ22+2cosθ}−{cosθ2+sin2θ2}](1+cosθ)2×2cosθ2+sin2θ2×2cosθ2−cosθ2−sin2θ2]checkuptothis...
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