integration of (1+sinx)/sinx(1+cosx) dx
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Int dx/(sinx+cosx)= Int (cosx-Sinx)dx/(Cos2x-sin2x)= Int cosx dx/ (1-2Sin2x) - Int sinx dx/ (2cos2x-1) ..(.1)
Let 1-2sin2x =u and 2cos2x-1 =v
so, du= -2 (2)cosx dx =-4cosx dx, thus cosx dx = -1/4 du........(2)
and dv= -2 (2) sinx dx = -4sin x dx, thus sinx dx= -1/4 dv....(3)
let's now plug-in the value of sin x and cosx in (1)
We have int -1/4 (du)/u - Int (-1/4) dv/v
=-1/4 ln(u) -1/4 ln (v)+C
= -1/4 ln (2cos2x-1)-1/4 ln (1-2sin2x) +C
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