Task/26882565
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7.
∫x*√(5x²+1)dx =(1/10)*∫√(5x²+1)d(5x²+1) =(1/10)*(2/3)*√(5x²+1)³ +C=
(1/15)*√(5x²+1)³ +C
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8.
∫(x+2) /(2x²+8x -7) dx =(1/4)*∫(1 / (2x²+8x -7) )* d(2x²+8x -7) =
(1/4)*Ln|2x²+8x -7| +C .
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9.
∫ x*Lg(x+2)*dx =(1/Ln10)*∫ x*Ln(x+2)*dx =(1/2Ln10)*∫ Ln(x+2)*d(x²) =
(1/2Ln10)*( x²*Ln(x+2)-∫ x²d(Ln(x+2)=(1/2Ln10)*( x²*Ln(x+2)-∫x²/(x+2)*dx )=
(1/2Ln10)*( x²*Ln(x+2) - x²/2 +2x - 4*Ln(x+2) ) + C .
* * * ∫ x²/(x+2) *dx = ∫ (x²- 4+4)/(x+2) *dx= ∫ (x-2)*dx + ∫ 4 /(x+2) *dx=
x²/2 -2x +4*Ln(x+2) * * *
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10.
∫ 1/ 9x²+25) +2/cos²(2x) )*dx =∫1/(9x²+25)*dx +∫2 /cos²(2x) *dx =
=(1/25)*∫(5/3)/(1 +(3x/5)²)*d(3x/5) +∫1/cos²(2x) *d(2x) =
(1/25)*(5/3)*∫1/(1 +(3x/5)²)*d(3x/5) +∫1/cos²(2x) *d(2x) =
(1/15)*arctg(3x/5) +tg(2x) +C .
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11.
∫sin²x*cosx*dx =∫sin²x*d(sinx) = (1/3)*sin³x +C.
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12.
∫cos(2x²+3)*x*dx =(1/4)*∫cos(2x²+3)*d(2x²+3) =(1/4)*sin(2x²+3)+C.
а)<span>sin(a)cos(b)*sin(a)cos(b)-sin(a)cos(b)*cos(a)sin(b)+cos(a)sin(b)*sin(a)cos(b)-cos(a)sin(b)*cos(a)sin(b)= </span>
<span>{-sin(a)cos(b)*cos(a)sin(b)+cos(a)sin(b)*sin(a)cos(b) сокращаются поскольку это одно и то же но с разными знаками и остается}</span>
<span>=sin^2a cos^2b-cos^2a sin^2b=</span>
<span>=sin^2a(1-sin^2b) - cos^2a sin^2b=</span>
<span>= sin^2a - sin^2b(sin^2a+cos^2a)=</span>
<span>= sin^2a - sin^2b.</span>
<span>б)не смогла решить.</span>