Help with a line integral please

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SUMMARY

The discussion focuses on solving the line integral ∫zds=∫acos(t)*( (acos(2t))^2+(2asin(t))^2+(-asin(t))^2 )^1/2 dt for the interval (0≤t≤pi/2). The user initially struggled with the integration but identified an error in the derivative dy/dt, which should be 2a sin(t) cos(t) = a sin(2t). After correcting this, the user successfully solved the integral using the substitution u=sin(t).

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Homework Statement
Show that the curve C given by
r=(a*cost*sint)i+(a*sin^2(t))j+(a*cos(t))k
, (0≤t≤pi/2)
lies on a sphere centred at the origin.
Find ∫zds of C
Relevant Equations
∫ds=∫F(r(t))·r'(t)dt
∫zds=∫acos(t)*( (acos(2t))^2+(2asin(t))^2+(-asin(t))^2 )^1/2 dt , (0≤t≤pi/2)
Simplified :
∫a^2cos(t)*(cos^2(2t)+5sin^2(t) )^1/2 dt , (0≤t≤pi/2)
However here i get stuck and i can´t find a way to rewrite it better or to integrate as it is.
Can i please get some help in this?
 
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I think you have dy/dt incorrect; it should be 2a \sin t \cos t = a \sin (2t).
 
pasmith said:
I think you have dy/dt incorrect; it should be 2a \sin t \cos t = a \sin (2t).
Oh okay!
I manged to solve the integration with the change and with the substitution u=sin(t)
Thank you!
 

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