Integral of sin(7x−(pi/4)) dx
The solution
Detail solution
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Let u=7x−4π.
Then let du=7dx and substitute 7du:
∫49sin(u)du
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The integral of a constant times a function is the constant times the integral of the function:
∫7sin(u)du=7∫sin(u)du
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The integral of sine is negative cosine:
∫sin(u)du=−cos(u)
So, the result is: −7cos(u)
Now substitute u back in:
−7cos(7x−4π)
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Now simplify:
−7sin(7x+4π)
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Add the constant of integration:
−7sin(7x+4π)+constant
The answer is:
−7sin(7x+4π)+constant
The answer (Indefinite)
[src]
/ / pi\
| cos|7*x - --|
| / pi\ \ 4 /
| sin|7*x - --| dx = C - -------------
| \ 4 / 7
|
/
−7cos(7x−4π)
The graph
/ pi\
sin|7 + --| ___
\ 4 / \/ 2
- ----------- + -----
7 14
7cos(4π)−7cos(4π−28)
=
/ pi\
sin|7 + --| ___
\ 4 / \/ 2
- ----------- + -----
7 14
−7sin(4π+7)+142
Use the examples entering the upper and lower limits of integration.