Integral of x^2cos(3x^3+5) dx
The solution
Detail solution
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Let u=3x3+5.
Then let du=9x2dx and substitute 9du:
∫81cos(u)du
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The integral of a constant times a function is the constant times the integral of the function:
∫9cos(u)du=9∫cos(u)du
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The integral of cosine is sine:
∫cos(u)du=sin(u)
So, the result is: 9sin(u)
Now substitute u back in:
9sin(3x3+5)
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Now simplify:
9sin(3x3+5)
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Add the constant of integration:
9sin(3x3+5)+constant
The answer is:
9sin(3x3+5)+constant
The answer (Indefinite)
[src]
/
| / 3 \
| 2 / 3 \ sin\3*x + 5/
| x *cos\3*x + 5/ dx = C + -------------
| 9
/
∫x2cos(3x3+5)dx=C+9sin(3x3+5)
1 sin(8) 1 sin(8)
<- - - ------, - - ------>
9 9 9 9
⟨−91−9sin(8),91−9sin(8)⟩
=
1 sin(8) 1 sin(8)
<- - - ------, - - ------>
9 9 9 9
⟨−91−9sin(8),91−9sin(8)⟩
Use the examples entering the upper and lower limits of integration.