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intergal of 3x^5 sin(x^3)dx
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\[\int\limits_{}^{}3x^5 \sin(x^3) dx\]
i chose u= sin(x^3) dv= 3x^5
you should : let t = x^3 --> dt = 3x^2 dx and \(x^5 = x^2x^3\) \[\rightarrow \int t~sin~t~ dt\]
now, take it by parts by letting u =t and dv = sint dt
i got du= 3x^2 cos(x^3) dx
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v= 1/2 x^6
you got stuck when the exponent of x is bigger and bigger if you go on that way
oh no wonder it keep going forever
Try my way!! :)
happend to 3x^5?
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\(x^5 =x^{3+2}=x^3*x^2\)
oooooooooooooooo
what does oooooooooooooooo mean?
'oh' lol
got it now, right? :)
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let me try :D
so for 3 i bring it out side?
nope, let t = x^3 --> dt = 3x^2 dx "3" is used here--> no need to move it out
|dw:1408646542552:dw|
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