Integral of sin(3x)/sqrt(cos(3x)) 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=3x.
Then let du=3dx and substitute 3du:
∫3cos(u)sin(u)du
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The integral of a constant times a function is the constant times the integral of the function:
∫cos(u)sin(u)du=3∫cos(u)sin(u)du
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Let u=cos(u).
Then let du=−sin(u)du and substitute −du:
∫(−u1)du
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The integral of a constant times a function is the constant times the integral of the function:
∫u1du=−∫u1du
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The integral of un is n+1un+1 when n=−1:
∫u1du=2u
So, the result is: −2u
Now substitute u back in:
−2cos(u)
So, the result is: −32cos(u)
Now substitute u back in:
−32cos(3x)
Method #2
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Let u=cos(3x).
Then let du=−2cos(3x)3sin(3x)dx and substitute −32du:
∫(−32)du
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The integral of a constant times a function is the constant times the integral of the function:
-
The integral of a constant is the constant times the variable of integration:
∫1du=u
So, the result is: −32u
Now substitute u back in:
−32cos(3x)
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Add the constant of integration:
−32cos(3x)+constant
The answer is:
−32cos(3x)+constant
The answer (Indefinite)
[src]
/
| __________
| sin(3*x) 2*\/ cos(3*x)
| ------------ dx = C - --------------
| __________ 3
| \/ cos(3*x)
|
/
∫cos(3x)sin(3x)dx=C−32cos(3x)
The graph
________
2 2*\/ cos(3)
- - ------------
3 3
32−32cos(3)
=
________
2 2*\/ cos(3)
- - ------------
3 3
32−32cos(3)
(0.579862216335156 - 0.795858425143173j)
(0.579862216335156 - 0.795858425143173j)
Use the examples entering the upper and lower limits of integration.