Question

integral cotx/logsinx

Answer 1

$\mathcal{\huge{\purple{\star\:Hayy \: Mate...!! \:\star}}}$


$<i>$
Ans : I = ⌡ {cotx / log(sinx) } dx

Let u=log(sinx)

du/dx = (1/sinx) * d(sinx)/dx

du/dx = cosx/sinx

du = cotx dx

Therefore : I = ⌡ {1 / u } du

I = log(u) + C

I = log(logIsinxI)+C

$<b>$

$<Marquee>$❤️⭐I hope you mark as brainlist answer⭐❤️✨✨✨✨