Question

integral of sin²x-cos²x/sinxcosx

Answer 1

${sin}^{2} x - {cos}^{2} x = - ( {cos}^{2} x - {sin}^{2} x) = - cos2x \\ also \\ sinxcosx \times 2 \div 2 = (1 \div 2)sin2x \\ question \: becomes \: integraion \: of \: \\ - 2cos2xdx \div sin2x \\ let \: sin2x = t \\ 2cos2xdx = dt \\ question \: reduces \: to \: integration \: of \: \: \\ - dt \div t \\ = - log(t) = - log(sin2x)$
here is complete answer

Answer 2

answer : ln|secx| - ln|sinx| + K

we have to integrate (sin²x - cos²x)/sinx.cosx

first we should resolve into simpler form.

(sin²x - cos²x)/sinx.cosx

= sin²x/sinx.cosx - cos²x/sinx.cosx

= (sinx/cosx) - (cosx/tanx)

= tanx - cotx

now we have to integrate (tanx - cotx) in place of (sin²x - cos²x)/sinx.cosx

from application of Integration,

$\int{tanx}\,dx=ln|secx|+C$

and $\int{cotx}\,dx=ln|sinx|+C$

so, ∫(tanx - cotx)dx = ∫tanx dx - ∫cotx dx

= ln|secx| - ln|sinx| + K [ where k is constant ]

hence, solution is ln|secx| - ln|sinx| + K

it's easy, isn't ?

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