Integral of cos^3xsinx dx
The solution
Detail solution
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There are multiple ways to do this integral.
Method #1
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Let u=cos(x).
Then let du=−sin(x)dx and substitute −du:
∫u3du
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The integral of a constant times a function is the constant times the integral of the function:
∫(−u3)du=−∫u3du
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The integral of un is n+1un+1 when n=−1:
∫u3du=4u4
So, the result is: −4u4
Now substitute u back in:
−4cos4(x)
Method #2
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Rewrite the integrand:
sin(x)cos3(x)=(1−sin2(x))sin(x)cos(x)
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Let u=sin2(x).
Then let du=2sin(x)cos(x)dx and substitute du:
∫(21−2u)du
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Integrate term-by-term:
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The integral of a constant is the constant times the variable of integration:
∫21du=2u
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The integral of a constant times a function is the constant times the integral of the function:
∫(−2u)du=−2∫udu
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The integral of un is n+1un+1 when n=−1:
∫udu=2u2
So, the result is: −4u2
The result is: −4u2+2u
Now substitute u back in:
−4sin4(x)+2sin2(x)
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Add the constant of integration:
−4cos4(x)+constant
The answer is:
−4cos4(x)+constant
The answer (Indefinite)
[src]
/
| 4
| 3 cos (x)
| cos (x)*sin(x) dx = C - -------
| 4
/
−4cos4x
The graph
4
1 cos (1)
- - -------
4 4
41−4cos41
=
4
1 cos (1)
- - -------
4 4
−4cos4(1)+41
Use the examples entering the upper and lower limits of integration.