Question

I am doing another integral....
I'll need a verification, if I am wrong, or when I am done.

Answer 1

the problem is:\[\int\limits_{ }^{ }x \sin(x) \cos(x) dx\]

Answer 2

I am thinking of differentiating the x, and integrating sin(x)cos(x).

Answer 3

\[\int\limits_{ }^{ }\sin(x)\cos(x)~dx=-\sin^2(x)-\int\limits_{ }^{ }\cos^2(x)~dx\]

Answer 4

just a tip. use the identity sin2x = 2sinxcosx =))

Answer 5

oh, tnx, I'll definitely try that

Answer 6

you still have question? feel free to ask :)

Answer 7

\[\int\limits_{ }^{ }x \sin(x)\cos(x)~dx\]\[\frac{1}{2}\int\limits_{ }^{ }x \sin(2x)~dx\]

Answer 8

that's right

Answer 9

\[\frac{1}{2}\int\limits_{ }^{ }x \sin(2x)~dx=-\frac{1}{4}x \cos(2x)-\frac{1}{4}\int\limits_{ }^{ }\cos(2x)~dx\]

Answer 10

\[\frac{1}{2}\int\limits_{ }^{ }x \sin(2x)~dx=-\frac{1}{4}x \cos(2x)+\frac{1}{8}\int\limits_{ }^{ }\sin(2x)+C\]

Answer 11

the second term of your answer is incorrect.

Answer 12

I have 1/2 from a chain of (2x), and it becomes positive, because inetgral of sin(x) is -cos(x).

are you sure it's incorrect?

Answer 13

I mena integral of cos(x) is sin(x). you are right it is -1/8

Answer 14

the \(\frac18 \int \sin(2x)\,dx\) part

Answer 15

\[\frac{1}{4}x \cos(2x)- \frac{1}{8} \sin(2x)+C\]

Answer 16

great!

Answer 17

so it's right?

Answer 18

To make you confidence of your work, you can always take the derivative of your answer to verify that it is the same as the function that you integrate

Answer 19

well, when I take the derivative I get.
\[\frac{1}{4}\cos(2x)-\frac{1}{2}x \sin(2x)-\frac{1}{4}\cos(2x).\]\[-\frac{1}{2}x \sin(2x)\]\[-x \sin(x) \cos(x)\]

Answer 20

I am getting a negative value....

Answer 21

I went wrong on the derivative ...

Answer 22

the answer should be:\[-\frac{1}{4}\cos(2x)-\frac{1}{8}\sin(2x)+C\]

Answer 23

my bad, it's\[-\frac{1}{4}\cos(2x)-\frac{1}{8}x \sin(2x)+C\]

Answer 24

now I will finally take the derivaitve.

Answer 25

now, I am lost, I got the integral incorrectly, most likely.

Answer 26

\(-\frac{1}{4}x\cos(2x)+\frac{1}{8}\sin(2x)+C\)

Answer 27

wait it is positive 1/8 sin(2x) ?

Answer 28

\[-\frac{1}{4}\cos(2x)+\frac{1}{8}\sin(2x)+C\]derivative.\[\frac{1}{2}x \sin(2x)-\frac{1}{4}\cos(2x)+\frac{1}{4}\cos(2x) \]

Answer 29

which becomes,\[x \sin(x)\cos(x)\]