Integral of sin^5xcosxdx 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=sin(x).
Then let du=cos(x)dx and substitute du:
∫u5du
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The integral of un is n+1un+1 when n=−1:
∫u5du=6u6
Now substitute u back in:
6sin6(x)
Method #2
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Rewrite the integrand:
sin5(x)cos(x)1=(1−cos2(x))2sin(x)cos(x)
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Let u=1−cos2(x).
Then let du=2sin(x)cos(x)dx and substitute 2du:
∫4u2du
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The integral of a constant times a function is the constant times the integral of the function:
∫2u2du=2∫u2du
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The integral of un is n+1un+1 when n=−1:
∫u2du=3u3
So, the result is: 6u3
Now substitute u back in:
6(1−cos2(x))3
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Add the constant of integration:
6sin6(x)+constant
The answer is:
6sin6(x)+constant
The answer (Indefinite)
[src]
/
| 6
| 5 sin (x)
| sin (x)*cos(x)*1 dx = C + -------
| 6
/
6sin6x
The graph
6sin61
=
6sin6(1)
Use the examples entering the upper and lower limits of integration.