anonymous
  • anonymous
integrate (x^2-6)cosx dx
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
integrate by parts
anonymous
  • anonymous
Does u=(x^2-6)?
anonymous
  • anonymous
int((x^2-6)cosx dx) =int(x^2cosx -6cosx dx) =int(x^2cosx dx) - int(6cosx dx)

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anonymous
  • anonymous
Ohhh, so you distribute before you even do anything?
anonymous
  • anonymous
\[\int\limits \left( x ^{2}-6 \right)\cos x dx=-6\int\limits \cos x dx+\int\limits x ^{2}\cos x dx+c\]
anonymous
  • anonymous
=-6 sin x+I1+c solve I1 by parts
anonymous
  • anonymous
int(x^2cosx dx) =x^2 int(cosx) - int((d(x^2)/dx)*int(cosx) dx) = -x^2sinx - (-2)int(xsinx dx) =-x^2sinx + 2[x int(sinx)- int((d(x)/dx)*int(sinx) dx)] =-x^2sinx +2[x cosx- int(sinx dx)] =-x^2sinx+2xcosx-2cosx
anonymous
  • anonymous
Hm, still getting it wrong
anonymous
  • anonymous
\[\int\limits x ^{2}\cos x dx=x ^{2}\sin x-\int\limits2x \sin x dx\]
dumbcow
  • dumbcow
@paul1231 , could you break it down into steps that are more understandable, it may be hard to follow your work u = x^2 ......... dv = cos x du = 2x .......... v = sin x \[= x^{2} \sin x -2 \int\limits x \sin x dx\] repeat int by parts u = x ............ dv = sin x du = dx .......... v = -cos x \[\int\limits x \sin x = -x \cos x +\int\limits \cos x = -x \cos x + \sin x\] sub that back in \[\int\limits x^{2} \cos x = x^{2} \sin x +2x \cos x -2 \sin x\] add the other int (-6cos) = -6sin x \[\int\limits (x^{2}-6)\cos x = x^{2} \sin x +2x \cos x -8 \sin x +C\]
anonymous
  • anonymous
\[\int\limits 2x \sin x dx=2\left[ x \left( -\cos x \right)-\int\limits 1\left( -\cos x \right)dx \right]\]
anonymous
  • anonymous
I got it :) your last cos in the answer was supposed to be a sin
anonymous
  • anonymous
@dumbcow well you did it for me
anonymous
  • anonymous
thank you, :)

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