Question

integral of x*cos(x/2) dx is the question
How does the answer become
2xsin(x/2) +4cos(x/2)+c?
Shouldnät the answer be x^2/2*sin(x/2) +c?

Answer 1

How did you arrive at \(\frac{1}{2}x^2\sin\left(\frac{x}{2}\right)+c\)?

Answer 2

well i broke them into parts, so x became x^2/2
and cos(x/2) became sin(x/2) +c

Answer 3

I ment sin(x/4) not sin(x/2)

Answer 4

You cannot break them apart. Just like you have to use product rule for product of two functions in differentiation, you have to use integration by parts for these kind of situation.

Answer 5

oh okey, but If I write them together, how should I begin?

Answer 6

This question is solved from integration by parts.
Let x=u and cos(x/2)=dv
Then dx=du and 2sin(x/2)=v
From the forumla we need to find u*v- integral(v*du)
=x*2sin(x/2)-integral(2sin(x/2)dx)
=2x.sin(x/2)+4cos(x/2)+c

Answer 7

oh okey I was just trying that., but instead of making cos(x/2)=dv I made it as v and then derrivated. Another thing, how can I know which derrivate law is the right one for every question?

Answer 8

The importance order for them
Logharitma, Arc (arctan arcsin etc.), Polynomial (x x^2 etc.), Trigonomethrical (sinx cosx) and Exponential numbers (2^x etc.).
You need to determine u with respect to this order.

Answer 9

for example, \[\int\limits_{}^{}a^x \] where a is constant and x is variable.

Answer 10

its logarithmic, but how do I go from here?

Answer 11

For example the derivative of 2^x is 2^x*ln2
so the integral of 2^x must be 2^x/ln2.
You dont have to apply integration by parts here.

Answer 12

oh now I see, thanx.

Answer 13

You're welcome.