Integral of t^3sin(2t^4)dt dx
The solution
Detail solution
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Let u=2t4.
Then let du=8t3dt and substitute 8du:
∫64sin(u)du
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The integral of a constant times a function is the constant times the integral of the function:
∫8sin(u)du=8∫sin(u)du
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The integral of sine is negative cosine:
∫sin(u)du=−cos(u)
So, the result is: −8cos(u)
Now substitute u back in:
−8cos(2t4)
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Add the constant of integration:
−8cos(2t4)+constant
The answer is:
−8cos(2t4)+constant
The answer (Indefinite)
[src]
/
| / 4\
| 3 / 4\ cos\2*t /
| t *sin\2*t /*1 dt = C - ---------
| 8
/
−8cos(2t4)
The graph
81−8cos2
=
−8cos(2)+81
Use the examples entering the upper and lower limits of integration.