## Integral Question

1. The problem statement, all variables and given/known data

Integrate[Cos[(2x+3)^(1/3)]

2. Relevant equations

3. The attempt at a solution

Would I use simple substitution?

Would it just simply be 1/2*Sin((2x+3)^(1/3))??
 PhysOrg.com science news on PhysOrg.com >> New language discovery reveals linguistic insights>> US official: Solar plane to help ground energy use (Update)>> Four microphones, computer algorithm enough to produce 3-D model of simple, convex room
 Recognitions: Homework Help 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?
 Your proposed solution completely disregards the 1/3 power. I would suggest integration by parts.

## Integral Question

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] ???
 Recognitions: Homework Help The substitution is correct, but your du is not. The expression should be multiplied by 2/3, not divided. So we have the following: $$du=\frac{2}{3} \left(\frac{1}{(2x+3)^{\frac{1}{3}}}\right)^2 dx$$ Now write the bracket expression in terms of u.

 Tags cos, integration, sin

 Similar Threads for: Integral Question Thread Forum Replies Calculus & Beyond Homework 0 Calculus & Beyond Homework 3 Calculus & Beyond Homework 1 Calculus & Beyond Homework 2 Calculus & Beyond Homework 2