Verified answer

1.) ∫sin^4(6θ) dθ = ∫sin^2(6θ)*sin^2(6θ) dθ

= 1/4*∫(1-cos(12θ))^2 dθ

= 1/4*∫(1- 2cos(12θ) + 1/2*(1+cos(24θ )) dθ

= 3θ/8 - 1/24*sin(12θ) + 1/(192)*sin(24θ) + C

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2.) ∫tan^3(2t)*sec^3(2t) dt

= ∫sec(2t)*tan(2t)*tan^2(2t)*sec^2(2t) dt

∫sec(2t)*tan(2t)*(sec^2(2t)-1)*sec^2(2t) dt

z = sec(2t)

dz/2 = sec(2t)*tan(2t) dt

1/2*∫(z^2 - 1)*z^2 dz

1/2*[(z^5)/5 - (z^3)/3] + C

sec^3(2t)/2*(sec^2(2t)/5 - 1/3) + C

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3.) ∫e^(x)*√(1-e^(2x)) dx

z = e^(x) => dz=e^(x) dx

∫√(1-z^2) dz

Let z=sin(θ) => dz=cos(θ) dθ

∫cos^2(θ) dθ = 1/2*(θ + sin(θ)cos(θ)) + C

arcsin(z)/2 + z√(1-z^2) + C

arcsin(e^(x))/2 + e^(x)*√(1-e^(2x)) + C

a million. ? (a million+x)²/?(x) dx For the integrand (x+a million)²/?(x), exchange u = ?(x) and du = a million/(2 ?(x)) dx: = 2 ? (u²+a million)² du For the integrand (u²+a million)², do long branch: = 2 ? (u^4+2 u²+a million) du combine the sum term by using term and factor out constants: = 2 ? u^4 du+4 ? u² du+2 ? a million du The ? of u² is u³/3: = 2 ? u^4 du+(4 u³)/3+2 ? a million du The ? of u^4 is u^5/5: = (2 u^5)/5+(4 u³)/3+2 ? a million du The ? of a million is u: = (2 u^5)/5+(4 u³)/3+2 u + C exchange back for u = ?(x): = (2 x^(5/2))/5+(4 x^(3/2))/3+2 ?(x) + C that's comparable to: = 2/15 ?(x) (3 x²+10 x+15) + C 2. ? x^6/(a million+x²) dx For the integrand x^6/(x²+a million), do long branch: = ? (x^4-x²-a million/(x²+a million)+a million) dx combine the sum term by using term and factor out constants: = ? x^4 dx- ? x² dx- ? a million/(x²+a million) dx+ ? a million dx The ? of a million/(x²+a million) is arctan(x): = ? x^4 dx- ? x² dx-arctan(x)+ ? a million dx The ? of x² is x³/3: = ? x^4 dx-x³/3-arctan(x)+ ? a million dx The ? of x^4 is x^5/5: = x^5/5-x³/3-arctan(x)+ ? a million dx The ? of a million is x: = x^5/5-x³/3+x-arctan(x) + C desire this enables! :)