John Marsh
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- May 9, 2013
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Hello Everyone please help me with this question
Find` int_(pi/2)^(-pi/2)` x sinx dx
Thanks in advance
Find` int_(pi/2)^(-pi/2)` x sinx dx
Thanks in advance
Hello Everyone please help me with this question
Find` int_(pi/2)^(-pi/2)` x sinx dx
Thanks in advance
Integration by parts. Let u= x, dv= sin(x) dx.Hello Everyone please help me with this question
Find` int_(pi/2)^(-pi/2)` x sinx dx
Thanks in advance
\(\displaystyle \int(\pi/2)^{-\pi/2} x sinx dx\) - Let's make it more readable :cool:
Integration by parts. Let u= x, dv= sin(x) dx.
Do you not know the "integration by parts" formula?
\(\displaystyle \int u dv= uv- \int v du\)
Yes, if you let u= x, dv= sin(x)dx then du= dx and v= -cos(x).
Put those into the formula.
\(\displaystyle u = x\)
\(\displaystyle dv = \sin(x)\)
\(\displaystyle v = -\cos(x)\)
\(\displaystyle du = 1\)
\(\displaystyle \int x \sin(x) = x -\cos(x) - \int-\cos(x)(1)\)
\(\displaystyle \int x \sin(x) = x -\cos(x) - \sin(x)x\) - Is this line right?
No...
\(\displaystyle \int x \sin(x) dx = x* [-\cos(x)] - \int-\cos(x)(1)dx \ = \ -x * cos(x) + sin(x) + C\)
Do you not know the difference between multiplying and adding?\(\displaystyle u = x\)
\(\displaystyle dv = \sin(x)\)
\(\displaystyle v = -\cos(x)\)
\(\displaystyle du = 1\)
\(\displaystyle \int x \sin(x) = x -\cos(x) - \int-\cos(x)(1)\)
\(\displaystyle \int x \sin(x) = x -\cos(x) - \sin(x)x\) - Is this line right?
Do you not know the difference between multiplying and adding?
You want x times -cos(x), not x plus -cos(x).
\(\displaystyle \displaystyle \int_{\frac{\pi}{2}}^{-\frac{\pi}{2}}x \ \ sin(x) \ \ dx\)
First solve the integral.
Use the "Integration by Parts" formula (because we are dealing with a product in the original problem):
\(\displaystyle \int u dv= uv- \int v du\)
\(\displaystyle u = x\)
\(\displaystyle du = 1\)
\(\displaystyle v = -\cos(x)\)
\(\displaystyle dv = \sin(x) \)
\(\displaystyle \int x \sin(x) = [x (-\cos(x))] - \int [- cos(x)(1)]\)
\(\displaystyle \int x \sin(x) = -x \cos(x) - \int - cos(x)\)
\(\displaystyle \int x \sin(x) = -x \cos(x) - [- \sin(x)] + C\) - Maybe better understood with this missing step.
\(\displaystyle \int x \sin(x) = -x \cos(x) + \sin(x) + C \)
\(\displaystyle \int x \sin(x) = -x \cos(x) + \sin(x) + C \)
Now moving on to evaluating the integral at two different values (definite integral).