Question

integrate (x^2-6)cosx dx

Answer 1

integrate by parts

Answer 2

Does u=(x^2-6)?

Answer 3

int((x^2-6)cosx dx)
=int(x^2cosx -6cosx dx)
=int(x^2cosx dx) - int(6cosx dx)

Answer 4

Ohhh, so you distribute before you even do anything?

Answer 5

\[\int\limits \left( x ^{2}-6 \right)\cos x dx=-6\int\limits \cos x dx+\int\limits x ^{2}\cos x dx+c\]

Answer 6

=-6 sin x+I1+c
solve I1 by parts

Answer 7

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

Answer 8

Hm, still getting it wrong

Answer 9

\[\int\limits x ^{2}\cos x dx=x ^{2}\sin x-\int\limits2x \sin x dx\]

Answer 10

@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
\[\int\limits (x^{2}-6)\cos x = x^{2} \sin x +2x \cos x -8 \sin x +C\]

Answer 11

\[\int\limits 2x \sin x dx=2\left[ x \left( -\cos x \right)-\int\limits 1\left( -\cos x \right)dx \right]\]

Answer 12

I got it :) your last cos in the answer was supposed to be a sin

Answer 13

@dumbcow well you did it for me

Answer 14

thank you, :)