Integral of sinx/(2cosx)/2 dx
The solution
Detail solution
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The integral of a constant times a function is the constant times the integral of the function:
∫2⋅2cos(x)sin(x)dx=2∫2cos(x)sin(x)dx
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Let u=2cos(x).
Then let du=−2sin(x)dx and substitute −2du:
∫4u1du
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The integral of a constant times a function is the constant times the integral of the function:
∫(−2u1)du=−2∫u1du
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The integral of u1 is log(u).
So, the result is: −2log(u)
Now substitute u back in:
−2log(2cos(x))
So, the result is: −4log(2cos(x))
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Add the constant of integration:
−4log(2cos(x))+constant
The answer is:
−4log(2cos(x))+constant
The answer (Indefinite)
[src]
/
|
| sin(x) log(2*cos(x))
| ---------- dx = C - -------------
| 2*cos(x)*2 4
|
/
−4logcosx
The graph
-log(cos(1))
-------------
4
−4logcos1
=
-log(cos(1))
-------------
4
−4log(cos(1))
Use the examples entering the upper and lower limits of integration.