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Prove $$ \int_0^\theta\frac{\sin\theta\cos x}{(1-\cos\theta\cos x)^2}dx=\csc^2\theta+\frac{\pi}{2}\cot\theta\csc\theta $$

$$ \int_0^\theta\frac{\sin\theta\cos x}{(1-\cos\theta\cos x)^2}dx=-\frac{\sin\theta}{\cos\theta}\int_0^\theta\frac{-\cos\theta\cos x+1-1}{(1-\cos\theta\cos x)^2}dx\\ =-\tan\theta\int_0^\theta\bigg[\frac{1}{1-\cos\theta\cos x}-\frac{1}{(1-\cos\theta\cos x)^2}\bigg]dx\\ $$

How do I solve it ?. Can I use Leibniz rule here ?

Thanks @Peter Foreman

$$ \int_0^\theta\frac{\sin\theta\cos x}{(1-\cos\theta\cos x)^2}dx=-\tan\theta\int_0^\theta\Bigg[\frac{1}{1-\cos\theta.\dfrac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}}-\frac{1}{\bigg(1-\cos\theta.\dfrac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}\bigg)^2}\Bigg]dx\\ =-\tan\theta\int_0^\theta\Bigg[\frac{\sec^2\frac{x}{2}}{1-\cos\theta+\tan^2\frac{x}{2}[1+\cos\theta]}+\frac{(1+\tan^2\frac{x}{2})\sec^2\frac{x}{2}}{\bigg(1-\cos\theta+\tan^2\frac{x}{2}[1+\cos\theta]\bigg)^2}\Bigg]dx $$ Set $t=\tan\frac{x}{2}\implies dt=\frac{1}{2}\sec^2\frac{x}{2}dx$ $$ I_1=-\tan\theta\int_0^{\tan\frac{\theta}{2}}\frac{2dt}{1+t^2-\cos\theta[1-t^2]} $$

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  • $\begingroup$ Do you know the tangent half angle substitution? $\endgroup$ – Peter Foreman Jan 22 '20 at 15:06
  • $\begingroup$ u mean $\cos x=\frac{1-\tan^2x/2}{1+\tan^2x/2}$ ?.will it help ? $\endgroup$ – ss1729 Jan 22 '20 at 15:08
  • $\begingroup$ Yes. This integral then has an elementary antiderivative. $\endgroup$ – Peter Foreman Jan 22 '20 at 15:09
  • $\begingroup$ @PeterForeman I tried to include that, am I stuck again ?. Could you please have a look into it ? $\endgroup$ – ss1729 Jan 22 '20 at 15:36
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Let

$$I(\theta) = \int_0^\theta\frac{dx}{1-\cos\theta\cos x}$$

Take the derivative with respect to $\theta$ and express the given integral as

$$\int_0^\theta\frac{\sin\theta\cos x}{(1-\cos\theta\cos x)^2}dx=\csc^2\theta-I'(\theta)\tag 1$$

Use $\cos x = \frac{1-\tan\frac x2}{1+\tan\frac x2}$ and integrate $I(\theta) $ as follows,

$$I(\theta) = \int_0^\theta\frac{dx}{1-\cos\theta\cos x} =\int_0^\theta\frac{2d(\tan\frac x2)}{(1+\cos\theta)\tan^2\frac x2 + (1-\cos \theta)} $$ $$=\frac{2\tan\frac {\theta}2}{1-\cos\theta}\int_0^\theta\frac{d\left(\frac{\tan\frac x2}{\tan\frac \theta2}\right)}{\left(\frac{\tan\frac x2}{\tan\frac \theta2}\right)^2 + 1} =2\csc\theta\tan^{-1} \left(\frac{\tan\frac x2}{\tan\frac \theta2}\right)\bigg|_0^\theta =\frac\pi2 \csc\theta$$

Then, evaluate $I'(\theta) = -\frac\pi2\cot\theta\csc\theta$ and plug into (1) to obtain $$ \int_0^\theta\frac{\sin\theta\cos x}{(1-\cos\theta\cos x)^2}dx=\csc^2\theta+\frac{\pi}{2}\cot\theta\csc\theta $$

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  • $\begingroup$ Set $t=\tan\frac{x}{2}\implies dt=\frac{1}{2}\sec^2\frac{x}{2}dx$ $$ I(\theta)=\int_0^\theta\frac{\sec^2\frac{x}{2}dx}{(1+\cos\theta)\tan^2\frac{x}{2}+1-\cos\theta}=\int_0^\theta\frac{\sec^2\frac{x}{2}dx}{2\cos^2\frac{\theta}{2}\tan^2\frac{x}{2}+2\sin^2\frac{\theta}{2}}\\ =\int\frac{2dt}{2\cos^2\frac{\theta}{2}.t^2+2\sin^2\frac{\theta}{2}}=\frac{1}{\sin^2\frac{\theta}{2}}\int\frac{dt}{\Big(t\cot\frac{\theta}{2}\Big)^2+1}\\ $$ $\endgroup$ – ss1729 Jan 22 '20 at 18:37
  • $\begingroup$ $$ =\frac{1}{\sin^2\frac{\theta}{2}}.\tan\frac{\theta}{2}\int\frac{\cot\frac{\theta}{2}.dt}{\Big(t\cot\frac{\theta}{2}\Big)^2+1}=\frac{1}{\sin\frac{\theta}{2}\cos\frac{\theta}{2}}\bigg[\tan^{-1}\Big[t.\cot\frac{\theta}{2}\Big]\bigg]=\frac{\pi}{4}\csc x.\sec x $$ Sorry did I make some mistake somewhr ?. Could you please help ? $\endgroup$ – ss1729 Jan 22 '20 at 18:37
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    $\begingroup$ @ss1729 - what you derived is fine, except note that the result $\frac\pi4 \csc \frac {\theta}2\sec\frac {\theta}2 = \frac\pi2\cdot \frac1{2\sin\frac{\theta}2\cos\frac {\theta}2} = \frac\pi2\frac1{\sin \theta}= \frac\pi2\csc\theta$ $\endgroup$ – Quanto Jan 22 '20 at 19:45
  • $\begingroup$ ohh I missed that in the half angle. I am sorry can't believe I didn't see that. $\endgroup$ – ss1729 Jan 22 '20 at 19:51

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