SteveBell Messages 4 Reaction score 0 Thread starter Jun 12, 2009 #1 Homework Statement Integrate[Cos[(2x+3)^(1/3)] Homework Equations The Attempt at a Solution Would I use simple substitution? Would it just simply be 1/2*Sin((2x+3)^(1/3))??
Homework Statement Integrate[Cos[(2x+3)^(1/3)] Homework Equations The Attempt at a Solution Would I use simple substitution? Would it just simply be 1/2*Sin((2x+3)^(1/3))??
Cyosis Homework Helper Messages 1,495 Reaction score 5 Jun 12, 2009 #2 You will have to use a substitution yes. No your answer is wrong. When you're integrating always differentiate your final answer to see if it gives the correct result. That said do you have any ideas about the kind of substitution you want to use?
You will have to use a substitution yes. No your answer is wrong. When you're integrating always differentiate your final answer to see if it gives the correct result. That said do you have any ideas about the kind of substitution you want to use?
blerg Messages 64 Reaction score 0 Jun 12, 2009 #3 Your proposed solution completely disregards the 1/3 power. I would suggest integration by parts.
SteveBell Messages 4 Reaction score 0 Jun 12, 2009 #4 Indeed. U=(2x+3)^1/3 Du= ((2x+3)^(-2/3))/(2/3) dv=Cos(X) V= Sin(x) then use U*du=d*v-integral[du*v] ?
Indeed. U=(2x+3)^1/3 Du= ((2x+3)^(-2/3))/(2/3) dv=Cos(X) V= Sin(x) then use U*du=d*v-integral[du*v] ?
Cyosis Homework Helper Messages 1,495 Reaction score 5 Jun 12, 2009 #5 The substitution is correct, but your du is not. The expression should be multiplied by 2/3, not divided. So we have the following: [tex] du=\frac{2}{3} \left(\frac{1}{(2x+3)^{\frac{1}{3}}}\right)^2 dx[/tex] Now write the bracket expression in terms of u.
The substitution is correct, but your du is not. The expression should be multiplied by 2/3, not divided. So we have the following: [tex] du=\frac{2}{3} \left(\frac{1}{(2x+3)^{\frac{1}{3}}}\right)^2 dx[/tex] Now write the bracket expression in terms of u.