Integral of ln(sin(x))/sqrt(x) dx
The solution
Detail solution
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Use integration by parts:
∫udv=uv−∫vdu
Let u(x)=log(sin(x)) and let dv(x)=x1.
Then du(x)=sin(x)cos(x).
To find v(x):
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The integral of xn is n+1xn+1 when n=−1:
∫x1dx=2x
Now evaluate the sub-integral.
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The integral of a constant times a function is the constant times the integral of the function:
∫sin(x)2xcos(x)dx=2∫sin(x)xcos(x)dx
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Don't know the steps in finding this integral.
But the integral is
∫sin(x)xcos(x)dx
So, the result is: 2∫sin(x)xcos(x)dx
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Now simplify:
2xlog(sin(x))−2∫tan(x)xdx
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Add the constant of integration:
2xlog(sin(x))−2∫tan(x)xdx+constant
The answer is:
2xlog(sin(x))−2∫tan(x)xdx+constant
The answer (Indefinite)
[src]
/
/ |
| | ___
| log(sin(x)) | \/ x *cos(x) ___
| ----------- dx = C - 2* | ------------ dx + 2*\/ x *log(sin(x))
| ___ | sin(x)
| \/ x |
| /
/
3sin2(2x)+3cos2(2x)−6cos(2x)+3(3sin2(2x)+3cos2(2x)−6cos(2x)+3)∫6e2logxsin2(4x)−24e2logxsin(2x)sin(4x)+6e2logxcos2(4x)+(12e2logx−24e2logxcos(2x))cos(4x)+24e2logxsin2(2x)+24e2logxcos2(2x)−24e2logxcos(2x)+6e2logx((40x2−12πx+6log2)sin(2x)+((6−24log2)x−3π)cos(2x)−6x+3π)sin(4x)+(((24log2−6)x+3π)sin(2x)+(40x2−12πx+6log2)cos(2x)−6log2)cos(4x)+(−80x2+24πx−12log2)sin2(2x)+((24log2+6)x−3π)sin(2x)+(−80x2+24πx−12log2)cos2(2x)+(40x2−12πx+18log2)cos(2x)−6log2dx+x((3sin2(2x)+3cos2(2x)−6cos(2x)+3)log(sin2x+cos2x+2cosx+1)+(3sin2(2x)+3cos2(2x)−6cos(2x)+3)log(sin2x+cos2x−2cosx+1)−6log2sin2(2x)+(3π−10x)sin(2x)−6log2cos2(2x)+6log2cos(2x))
pi
--
2
/
|
| log(sin(x))
| ----------- dx
| ___
| \/ x
|
/
0
0∫2πxlog(sin(x))dx
=
pi
--
2
/
|
| log(sin(x))
| ----------- dx
| ___
| \/ x
|
/
0
0∫2πxlog(sin(x))dx
Use the examples entering the upper and lower limits of integration.