Question

$\huge \int \: \frac{ \cot(x) }{ log( \sin(x) ) }$
Solve it .....​

Answer 1

$\huge \red{\mathfrak {solution}}$

$\int \: \frac{ \cot(x) }{ log( \sin(x) ) }$

Here we use substitution method :

let log(sin(x))= t

=> 1/sinx(cosx)dx= dt

=> cosx/sinx dx = dt

=>. cotx = dt

now put these in question we get

$\implies \: \int \: \frac{dt}{t} \: \\ \\ \\ \implies \: log(t)$

Now put the value of t= log(sin(x))

$\implies \: log( log \sin(x) )$

This is the required answer ..

Answer 2

I = ⌡ {cotx / log(sinx) } dx

Let u = log(sinx)

On differentiating, we will get

du/dx = (1/sinx) * d(sinx)/dx [log x = (1/x)]

du/dx = cosx/sinx

du = cotx dx

Hence,

I = ⌡ {1 / u } du

=> I = log(u) + C

=> I = log(logIsinxI)+C

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