- #1
yungman
- 5,718
- 241
I don't understand why I solve the integration in two different ways and get two different answers!
To find:
[tex]\int_0^{\pi} P_1^1(cos \theta) sin \theta d \theta [/tex]
1) Solve in [itex] \theta[/itex]
[tex] P_1(cos \theta) = cos \theta \;\Rightarrow \; P_1^1(cos \theta)= -sin \theta [/tex]
[tex]\int_0^{\pi} P_1^1(cos \theta) sin \theta d \theta = -\int_0^{\pi}sin^2 \theta d \theta = -\frac{\pi}{2}[/tex]
2) Let [tex]s=cos \theta \;\Rightarrow \; d\theta = \frac{ds}{-sin \theta} [/tex]
[tex]\int_0^{\pi} P_1^1(cos \theta) sin \theta d \theta = -\int _1^{-1} P_1^1(s)ds[/tex]
[tex]P_1(s)=s \;\Rightarrow P_1^1(s)=1[/tex]
[tex]\int_0^{\pi} P_1^1(cos \theta) sin \theta d \theta = -\int _1^{-1} P_1^1(s)ds = -s|_1^{-1} = 2[/tex]
You see the two method yield two answers! I know it should yield the same answer, the book show how to solve the problems in 2) form. Please tell me what did I do wrong.
thanks
Alan
To find:
[tex]\int_0^{\pi} P_1^1(cos \theta) sin \theta d \theta [/tex]
1) Solve in [itex] \theta[/itex]
[tex] P_1(cos \theta) = cos \theta \;\Rightarrow \; P_1^1(cos \theta)= -sin \theta [/tex]
[tex]\int_0^{\pi} P_1^1(cos \theta) sin \theta d \theta = -\int_0^{\pi}sin^2 \theta d \theta = -\frac{\pi}{2}[/tex]
2) Let [tex]s=cos \theta \;\Rightarrow \; d\theta = \frac{ds}{-sin \theta} [/tex]
[tex]\int_0^{\pi} P_1^1(cos \theta) sin \theta d \theta = -\int _1^{-1} P_1^1(s)ds[/tex]
[tex]P_1(s)=s \;\Rightarrow P_1^1(s)=1[/tex]
[tex]\int_0^{\pi} P_1^1(cos \theta) sin \theta d \theta = -\int _1^{-1} P_1^1(s)ds = -s|_1^{-1} = 2[/tex]
You see the two method yield two answers! I know it should yield the same answer, the book show how to solve the problems in 2) form. Please tell me what did I do wrong.
thanks
Alan