Integration of 1-cotx/1+cotx. dx

1.	Integration of 1-cotx/1+cotx. dx
Answer» write cotx as cosx/sinx U GET 1-(cos/sinx)divided by 1+(cosx/sinx) u get (sinx-cosx)/(sinx+cosx) substitute sinx+cosx=u differentiate both sides u get (cosx-sinx).dx=du -(-cosx+sin X)dx=du -cosx+sinx dx= -du sinx-cosx dx=-du substitute BACK in the integral expression u get integral of(-du/u) = -log u substitute value of u

Answer»

write cotx as cosx/sinx

U GET 1-(cos/sinx)divided by 1+(cosx/sinx)

u get (sinx-cosx)/(sinx+cosx)

substitute sinx+cosx=u

differentiate both sides

u get (cosx-sinx).dx=du

-(-cosx+sin X)dx=du

-cosx+sinx dx= -du

sinx-cosx dx=-du

substitute BACK in the integral expression

u get integral of(-du/u)

= -log u

substitute value of u

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Integration of 1-cotx/1+cotx. dx

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