help with a calculus question

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help with a calculus question

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Int dx/(sinx+cosx)= Int (cosx-Sinx)dx/(Cos^{2}x-sin^{2}x)= Int cosx dx/ (1-2Sin^{2}x) - Int sinx dx/ (2cos^{2}x-1) ..(.1)

Let 1-2sin^{2}x =u and 2cos^{2}x-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 (2cos^{2}x-1)-1/4 ln (1-2sin^{2}x) +C

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

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

Hint: sin(x)+cos(x) = sqrt(2)sin(x+pi/4), and integral of 1/sin(x) = ln(|sin(x/2)| - ln(|cos(x/2)|

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