Question

Give the integration of 1/ Sin∅ + Cos ∅


























































































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Answer 1

Step-by-step explanation:

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 2

$\int \: \dfrac{1}{sin \: \theta \: . cos \: \theta} \: \: dx \\ \\ \implies \: \int \dfrac{sin {}^{2} \: \theta \: \: + cos {}^{2} \: \theta }{sin \: \theta.cos \: \theta} \: \: dx \\ \\ \\ \implies \: \int \: \dfrac{sin {}^{2} \: \theta }{ \: sin \: \theta.cos \: \theta} dx + \dfrac{cos {}^{2} \: \theta }{ \: sin \: \theta.cos \: \theta} \: dx$

<Separating the integral by dividing numerator and denominator >

$\implies \int \: tan \: \theta \: dx \: + \: \int \: cot \: \theta \: dx \\ \\ \implies \: log \: |sec \: \theta| + log \: \ |sin \: \theta| + c$