\[\large{\displaystyle \int^{\frac{\pi}{2}}_{0} x \ln(\cos x) \sqrt{\tan x} \text{ d}x}\]

plz post ur method and i dont know its answer

\[\large{\displaystyle \int^{\frac{\pi}{2}}_{0} x \ln(\cos x) \sqrt{\tan x} \text{ d}x}\]

plz post ur method and i dont know its answer

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TopNewestClosed form has been posted on math-stack exchange, I will post the link soon. – Ronak Agarwal · 1 year, 11 months ago

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your problem. – Kartik Sharma · 1 year, 11 months ago

I think that link will give a hint of how to doLog in to reply

Ok guys it was inspired from this – Tanishq Varshney · 1 year, 11 months ago

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@Tanishq Varshney This seems to involve Complex Contour Integration. Unfortunately I don't know Contour Integration. @Kartik Sharma Please could you help? – Ishan Dasgupta Samarendra · 1 year, 11 months ago

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– Kartik Sharma · 1 year, 11 months ago

I don't think it involves contour integration. It might be involving but then it is not that smooth to notice that easily.Log in to reply

@Krishna Sharma @Ishan Dasgupta Samarendra @Pi Han Goh – Tanishq Varshney · 1 year, 11 months ago

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– Ishan Dasgupta Samarendra · 1 year, 11 months ago

I'll try it though I don't know if I can get an answer.Log in to reply

this and I think it is essentially the same problem. But since I've not solved that problem, I can only help you until here. There maybe solvers of that problem out there and they can surely help. – Kartik Sharma · 1 year, 11 months ago

First look remind me of \(\frac{\beta'(a,b)}{4}\) at \(a = -1/4, b = -3/2\). But then there's that \(x\) which reminds me ofLog in to reply

– Pi Han Goh · 1 year, 11 months ago

Are you sure it has a closed form? If yes, what makes you say that?Log in to reply

Check this link – Ronak Agarwal · 1 year, 11 months ago

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– Tanishq Varshney · 1 year, 11 months ago

Thanx for the link , can u plz reply to the post where i commented on one of ur solutionLog in to reply

– Ronak Agarwal · 1 year, 11 months ago

where.Log in to reply

– Tanishq Varshney · 1 year, 11 months ago

Ok i'll mention u again there.Log in to reply