Question

Log(cot)xdx integration pie by 2 to 0

Answer 1

Answer:

I=integ.of (lim 0 to π/2) log(tan x+cot x).dx.

I=integ.of (lim 0 to π/2) log (sin^2x+cos^2x)/sin x.cos x. dx

I=integ.of (lim 0 toπ/2) log1/(sin x.cos x). dx

I= integ.of (lim 0 to π/2) (log 1 - log sin x.cos x). dx

I=integ.of (lim 0 to π/2)(0-log sin x. -log cos x ). dx

I=integ.of (lim 0 to π/2)[ - log sin x. - log cos x]

We know that :-

integral (0 to π/2) log sin x= integral (lim 0 to π/2) log cos x= - π/2.log2.

I= -(-π/2.log2). - (- π/2.log2)

I=π/2.log2 + π/2.log2

I=2(π/2.log 2)

I =π.log2. Proved.

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