- #1
yungman
- 5,755
- 293
Evaluate
[tex]\int_0^{\pi}\cos(a\sin\theta)d\theta[/tex]
This is my work so far and I don't know how to proceed or is this even correct
Let ##u=a\sin\theta \Rightarrow\; \frac {du}{a\cos\theta}=d\theta##
[tex]\int_0^{\pi}\cos(a\sin\theta)d\theta=\int\frac{\cos u}{a\cos\theta}du[/tex]
But I cannot set the limit as both are zero! So I just skip this step and put the interval back in after calculation.
Let ##W=a\cos\theta\Rightarrow\;dW=-a\sin\theta##. Let ##dv=\frac {1}{\cos\theta}=\sec\theta\Rightarrow\;v=\int\sec\theta d\theta=ln|\sec \theta+\tan\theta|##
This very complicated to substitute.
So I tried this
[tex]\int\frac{\cos u}{a\cos\theta}du=\frac {1}{a} \int \frac{\cos u}{\sqrt{1-\sin^2\theta}}du= \frac {1}{x} \int \frac{\cos u}{\sqrt{1-(\frac{u}{a})^2}}du [/tex]
Still it is very hard. Is that any better way?
Thanks
[tex]\int_0^{\pi}\cos(a\sin\theta)d\theta[/tex]
This is my work so far and I don't know how to proceed or is this even correct
Let ##u=a\sin\theta \Rightarrow\; \frac {du}{a\cos\theta}=d\theta##
[tex]\int_0^{\pi}\cos(a\sin\theta)d\theta=\int\frac{\cos u}{a\cos\theta}du[/tex]
But I cannot set the limit as both are zero! So I just skip this step and put the interval back in after calculation.
Let ##W=a\cos\theta\Rightarrow\;dW=-a\sin\theta##. Let ##dv=\frac {1}{\cos\theta}=\sec\theta\Rightarrow\;v=\int\sec\theta d\theta=ln|\sec \theta+\tan\theta|##
This very complicated to substitute.
So I tried this
[tex]\int\frac{\cos u}{a\cos\theta}du=\frac {1}{a} \int \frac{\cos u}{\sqrt{1-\sin^2\theta}}du= \frac {1}{x} \int \frac{\cos u}{\sqrt{1-(\frac{u}{a})^2}}du [/tex]
Still it is very hard. Is that any better way?
Thanks
Last edited: