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

Question

help with a calculus question

Answer 1

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)|

Answer 2

 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

 

Answer 3

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Answer 4

Using the trigonometric identities

cos x = sin (x+π/2) and sin x + sin y = 2 sin((x+y)/2) cos((x-y)/2) gives

1/(sin x + cos x) = 1/(sin x + sin(x+π/2)) = 1/(2sin(x+π/4)cos(-π/4)) = csc(x+π/4)/√2

We can integrate using the formula ∫ csc x dx = ln|tan(x/2)| + C to get

∫ dx/(sinx+cos x) = (1/√2) * ∫ [csc(x+π/4)] dx = (1/√2) ln|tan(x/2 + π/8)| + C