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1-let-f-a-a-tanx-dx-with-a-gt-0-find-a-explicit-form-for-f-x-2-find-also-g-a-dx-a-tanx-3-calculate-2-tanx-dx-and-dx-2-tanx-




Question Number 69044 by mathmax by abdo last updated on 18/Sep/19
1) let f(a) =∫(√(a+tanx))dx  with a>0  find a explicit form for f(x)  2) find also g(a) =∫  (dx/( (√(a+tanx))))  3)calculate ∫(√(2+tanx))dx and ∫  (dx/( (√(2+tanx))))
1)letf(a)=a+tanxdxwitha>0findaexplicitformforf(x)2)findalsog(a)=dxa+tanx3)calculate2+tanxdxanddx2+tanx
Answered by mind is power last updated on 18/Sep/19
u=(√(a+tgx))  x=arctg(u^2 −a)  dx=((2udu)/(1+(u^2 −a)^2 ))  ∫(√(a+tan(x)))dx=∫((2u^2 du)/((u^2 −a)^2 +1))  =∫((2u^2 du)/(u^4 −2au^2 +a^2 +1))=∫((2u^2 du)/((u^2 +(√(a^2 +1)))^2 −(2a+2(√(a^2 +1)))u^2 ))  =∫((2u^2 du)/((u^2 −(√((2a+2(√(a^2 +1)))))u+(√(a^2 +1)))(u^2 +(√(2a+(√(a^2 +1))))u+(√(a^2 +1)))))  β=(√(2a+2(√(a^2 +1))))  η=(√(a^2 +1))  ∫((2u^2 )/((u^2 −βx+η)(x^2 +βx+η)))=((ax+b)/(x^2 −βx+η))+((cx+d)/(x^2 +βx+η))  a+c=0  b+d=0  β(a−c)+d+b=2  η(c+a)+β(−d+b)=0  ⇒a=−c     b=−d  ,−2cβ=2  b=d   ,c=−(1/β),a=(1/β),b=d=0  ∫(x/(β(x^2 −βx+η)))dx−∫(dx/(β(x^2 +βx+η)))  =(1/β)∫(dx/((x−(β/2))^2 +η−(β^2 /4)))−(1/β)∫(dx/((x+(β/2))^2 +η−(β^2 /4)))  =(1/(β(√(η−(β^2 /4)))))tan^(−1) (((x−(β/2))/( (√(η−(β^2 /4))))))−(1/(β(√(η−(β^2 /4)))))tan^(−1) (((x+(β/2))/( (√(η−(β^2 /4))))))+c  ∫(√(a+tan(x)))dx=(1/(β(√(η−(β^2 /4)))))[tan^(−1) ((((√(a+tg(x)))−(β/2))/( (√(η−(β^2 /4))))))−tan^(−1) ((((√(a+tg(x)))+(β/2))/( (√(η−(β^2 /4))))))+c  f(a)=(1/( (√(2a+2(√(a^2 +1))(√(((√(a^2 +1))−a)/2))))))[tan^(−1) ((((√(a+tan (x)))−((√(2a+2(√(a^2 +1))))/2))/( (√(((√(a^2 +1))−a)/2)))))−tan^(−1) ((((√(a+tan(x)))+((√(2a+2(√(a^2 +1))))/2))/( (√(((√(a^2 +1))−a)/2)))))]  f′(a)=∫(d/da)((√(a+tan (x)))dx)=(1/2)∫(dx/( (√(a+tan(x)))))  ⇒∫(dx/( (√(a+tan(x)))))=2f′(a)′′verry long expression′′  3)∫(√(2+tan (x)))dx=f(2)  ∫(dx/( (√(2+tg(x)))))=2f′(2)
u=a+tgxx=arctg(u2a)dx=2udu1+(u2a)2a+tan(x)dx=2u2du(u2a)2+1=2u2duu42au2+a2+1=2u2du(u2+a2+1)2(2a+2a2+1)u2=2u2du(u2(2a+2a2+1)u+a2+1)(u2+2a+a2+1u+a2+1)β=2a+2a2+1η=a2+12u2(u2βx+η)(x2+βx+η)=ax+bx2βx+η+cx+dx2+βx+ηa+c=0b+d=0β(ac)+d+b=2η(c+a)+β(d+b)=0a=cb=d,2cβ=2b=d,c=1β,a=1β,b=d=0xβ(x2βx+η)dxdxβ(x2+βx+η)=1βdx(xβ2)2+ηβ241βdx(x+β2)2+ηβ24=1βηβ24tan1(xβ2ηβ24)1βηβ24tan1(x+β2ηβ24)+ca+tan(x)dx=1βηβ24[tan1(a+tg(x)β2ηβ24)tan1(a+tg(x)+β2ηβ24)+cf(a)=12a+2a2+1a2+1a2[tan1(a+tan(x)2a+2a2+12a2+1a2)tan1(a+tan(x)+2a+2a2+12a2+1a2)]f(a)=dda(a+tan(x)dx)=12dxa+tan(x)dxa+tan(x)=2f(a)verrylongexpression3)2+tan(x)dx=f(2)dx2+tg(x)=2f(2)
Commented by turbo msup by abdo last updated on 19/Sep/19
thank you sir.
thankyousir.
Commented by mind is power last updated on 19/Sep/19
y′re welcom
yrewelcom

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