Question

solve integral log(tan x+cot x) limits 0 to pie//2)

Answer 1

Log(tanx +cotx) 

= log (sinx/cosx/sinx) 

= log(sin squarex + cos square x / sinxcosx) 

= log(1/sinxcosx 

= log(1)-log(sinxcosx) 

= -[log(sinxcosx)] 

= -[log(sinx) + log(cosx)] 

You can integrate this by parts.

Answer 2

Intg ln[(sin^2x + cos^2x)/(sins.cosx)]

-lntg lnsinx+ -intg lncosx

Integration of lnsinx and lncosx both are among standard results for limit x=0 to pi/2 I.e. -pi/2 ×ln2

That will give u -(-pi/2 ×ln2) -(-pi/2 ×ln2)

=pi.ln2

that's the answer