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integral 1/(sin(x)+cos(x))

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3 Answers

 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

 

Comments

Asok, your u-substitution isn't correct. The derivative of cos2x -sin2x IS NOT -2sinx-2cosx. You didn't use the chain rule. The derivative of cos2x-sin2x is 2cosx(-sinx)-2sinxcosx = -4sinxcosx=-2sin2x.

You are still wrong. The derivative of 1-2sin2x IS NOT -4cosx. It is -4cosxsinx. You should stop trying to give mathematical advice because you don't know what you're talking about.