Integral of sqrt(2sin3x)cos3x dx
The solution
Detail solution
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Let u=3x.
Then let du=3dx and substitute 32du:
∫32sin(u)cos(u)du
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The integral of a constant times a function is the constant times the integral of the function:
∫sin(u)cos(u)du=32∫sin(u)cos(u)du
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Let u=sin(u).
Then let du=cos(u)du and substitute du:
∫udu
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The integral of un is n+1un+1 when n=−1:
∫udu=32u23
Now substitute u back in:
32sin23(u)
So, the result is: 922sin23(u)
Now substitute u back in:
922sin23(3x)
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Add the constant of integration:
922sin23(3x)+constant
The answer is:
922sin23(3x)+constant
The answer (Indefinite)
[src]
/
| ___ 3/2
| ____________ 2*\/ 2 *sin (3*x)
| \/ 2*sin(3*x) *cos(3*x) dx = C + -------------------
| 9
/
∫2sin(3x)cos(3x)dx=C+922sin23(3x)
The graph
___ 3/2
2*\/ 2 *sin (3)
-----------------
9
922sin23(3)
=
___ 3/2
2*\/ 2 *sin (3)
-----------------
9
922sin23(3)
Use the examples entering the upper and lower limits of integration.