Integral of 0,5*sin(x+y^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:
∫2sin(x+y2)dx=2∫sin(x+y2)dx
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Let u=x+y2.
Then let du=dx and substitute du:
∫sin(u)du
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The integral of sine is negative cosine:
∫sin(u)du=−cos(u)
Now substitute u back in:
−cos(x+y2)
So, the result is: −2cos(x+y2)
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Add the constant of integration:
−2cos(x+y2)+constant
The answer is:
−2cos(x+y2)+constant
The answer (Indefinite)
[src]
/
|
| / 2\ / 2\
| sin\x + y / cos\x + y /
| ----------- dx = C - -----------
| 2 2
|
/
∫2sin(x+y2)dx=C−2cos(x+y2)
/ 2\ / 2\
cos\y / cos\1 + y /
------- - -----------
2 2
2cos(y2)−2cos(y2+1)
=
/ 2\ / 2\
cos\y / cos\1 + y /
------- - -----------
2 2
2cos(y2)−2cos(y2+1)
cos(y^2)/2 - cos(1 + y^2)/2
Use the examples entering the upper and lower limits of integration.