yungman
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I don't understand why I solve the integration in two different ways and get two different answers!
To find:
\int_0^{\pi} P_1^1(cos \theta) sin \theta d \theta
1) Solve in \theta
P_1(cos \theta) = cos \theta \;\Rightarrow \; P_1^1(cos \theta)= -sin \theta
\int_0^{\pi} P_1^1(cos \theta) sin \theta d \theta = -\int_0^{\pi}sin^2 \theta d \theta = -\frac{\pi}{2}
2) Let s=cos \theta \;\Rightarrow \; d\theta = \frac{ds}{-sin \theta}
\int_0^{\pi} P_1^1(cos \theta) sin \theta d \theta = -\int _1^{-1} P_1^1(s)ds
P_1(s)=s \;\Rightarrow P_1^1(s)=1
\int_0^{\pi} P_1^1(cos \theta) sin \theta d \theta = -\int _1^{-1} P_1^1(s)ds = -s|_1^{-1} = 2
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:
\int_0^{\pi} P_1^1(cos \theta) sin \theta d \theta
1) Solve in \theta
P_1(cos \theta) = cos \theta \;\Rightarrow \; P_1^1(cos \theta)= -sin \theta
\int_0^{\pi} P_1^1(cos \theta) sin \theta d \theta = -\int_0^{\pi}sin^2 \theta d \theta = -\frac{\pi}{2}
2) Let s=cos \theta \;\Rightarrow \; d\theta = \frac{ds}{-sin \theta}
\int_0^{\pi} P_1^1(cos \theta) sin \theta d \theta = -\int _1^{-1} P_1^1(s)ds
P_1(s)=s \;\Rightarrow P_1^1(s)=1
\int_0^{\pi} P_1^1(cos \theta) sin \theta d \theta = -\int _1^{-1} P_1^1(s)ds = -s|_1^{-1} = 2
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