Need help with the integration of attached problem

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Shahryar
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Kindly can someone tell me how can i integrate and plot the following formula as a function of time. while theta is changing from 1 to 20.
also attached .docx file for convenience. Any kind of help or advice will be highly appreciable.
integration problem.jpg
 

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berkeman said:
Where is the problem from? Is it for schoolwork?

No its in my engineering design problem. I am just stuck here. I need to plot the graph for theta as a function of time. I am not really good in calculus to be honest.
May be i can use MATLAB to solve !
 
Shahryar said:
No its in my engineering design problem.

What does that mean? An engineering design project for what year in university? An engineering project for work? Where do you work?
 
It would strike me as an obvious first step to use the facts that [itex]sin(20- \theta)= sin(20)cos(\theta)- cos(20)sin(\theta)[/itex] and [itex]cos(20- \theta)= cos(20)cos(\theta)+ sin(20)sin(\theta)[/itex].

That "20" looks strange for radian measure. If it is in degrees you should remember that [itex]\int sin(x)dx= -cos(x)+ C[/itex] and [itex]\int cos(x)= sin(x)+ C[/itex] only for x measured in radians.
 
HallsofIvy said:
It would strike me as an obvious first step to use the facts that [itex]sin(20- \theta)= sin(20)cos(\theta)- cos(20)sin(\theta)[/itex] and [itex]cos(20- \theta)= cos(20)cos(\theta)+ sin(20)sin(\theta)[/itex].

That "20" looks strange for radian measure. If it is in degrees you should remember that [itex]\int sin(x)dx= -cos(x)+ C[/itex] and [itex]\int cos(x)= sin(x)+ C[/itex] only for x measured in radians.

Yes that 20 is actually 20 degrees. should i convert it into radians which is 0.34906585 radians ?