Jun 16, 2004 #1 jcsd Science Advisor Gold Member Messages 2,112 Reaction score 13 Can someone solve this integral as the answer I get looks suspicously complicated: \int^{0}_{\frac{-u}{a}} t\sqrt{1 - \frac{(u + at)^2}{c^2}} dt
Can someone solve this integral as the answer I get looks suspicously complicated: \int^{0}_{\frac{-u}{a}} t\sqrt{1 - \frac{(u + at)^2}{c^2}} dt
Jun 16, 2004 #2 AKG Science Advisor Homework Helper Messages 2,559 Reaction score 4 Make the substitution \frac{u + at}{c} = \cos \theta. You should be able to get it down to this: \frac{c}{a^2} \left (u\int _{\frac{\pi}{2}} ^{\arccos \left(\frac{u}{c}\right)} \sin ^2 \theta d\theta - c\int _{\frac{\pi}{2}} ^{\arccos \left(\frac{u}{c}\right)} \sin ^2 \theta \cos \theta d\theta \right ) And you can easily solve that on your own. EDITED to fix limits of integration as per HallsOfIvy's comment. Last edited: Jun 16, 2004
Make the substitution \frac{u + at}{c} = \cos \theta. You should be able to get it down to this: \frac{c}{a^2} \left (u\int _{\frac{\pi}{2}} ^{\arccos \left(\frac{u}{c}\right)} \sin ^2 \theta d\theta - c\int _{\frac{\pi}{2}} ^{\arccos \left(\frac{u}{c}\right)} \sin ^2 \theta \cos \theta d\theta \right ) And you can easily solve that on your own. EDITED to fix limits of integration as per HallsOfIvy's comment.
Jun 16, 2004 #3 HallsofIvy Science Advisor Homework Helper Messages 42,895 Reaction score 984 The substitution might work but the limits of integration are wrong. When t= 0, cos[theta]= u/c so [theta]= cos<sup>-1</sup>(u/c). When t= u/c, cos[theta]= 0 so [theta]= [pi]/2.
The substitution might work but the limits of integration are wrong. When t= 0, cos[theta]= u/c so [theta]= cos<sup>-1</sup>(u/c). When t= u/c, cos[theta]= 0 so [theta]= [pi]/2.
Jun 17, 2004 #4 jcsd Science Advisor Gold Member Messages 2,112 Reaction score 13 Thanks for that, I realized I made a slight error so it became slightly easier to solve.