Question

integration of sin x/sin x+cos x dx. ​

Answer 1

Step-by-step explanation:

Given ∫{sinx/(sinx+cosx)}dx

= (1/2)*∫{2sinx/(sinx+cosx)}dx

= (1/2)*∫{(2sinx+cosx-cosx)/(sinx+cosx)}dx

= (1/2)*∫{(sinx+sinx+cosx-cosx)/(sinx+cosx)}dx

= (1/2)*∫{(sinx+cosx +sinx-cosx)/(sinx+cosx)}dx

= (1/2)*∫{1 + (sinx-cosx)/(sinx+cosx)}dx

Let sinx + cosx = t

Differentiate with respect to x

(cosx - sinx)dx = dt

=> - (sinx - cosx)dx = dt

=> (sinx - cosx)dx = -dt

Now (1/2)*∫{1 + (sinx-cosx)/(sinx+cosx)}dx = -(1/2)*∫{1 + 1/t }dt

= -(1/2)*[x + logt] + c

= -(1/2)*[x + log(sinx + cosx)] + c (By putting sinx + cosx = t)

So ∫{sinx/(sinx+cosx)}dx = -(1/2)*[x + log(sinx + cosx)] + c

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Answer 2

Answer:

Given ∫{sinx/(sinx+cosx)}dx

= (1/2)*∫{2sinx/(sinx+cosx)}dx

= (1/2)*∫{(2sinx+cosx-cosx)/(sinx+cosx)}dx

= (1/2)*∫{(sinx+sinx+cosx-cosx)/(sinx+cosx)}dx

= (1/2)*∫{(sinx+cosx +sinx-cosx)/(sinx+cosx)}dx

= (1/2)*∫{1 + (sinx-cosx)/(sinx+cosx)}dx

Let sinx + cosx = t

Differentiate with respect to x

(cosx - sinx)dx = dt

=> - (sinx - cosx)dx = dt

=> (sinx - cosx)dx = -dt

Now (1/2)*∫{1 + (sinx-cosx)/(sinx+cosx)}dx = -(1/2)*∫{1 + 1/t }dt

= -(1/2)*[x + logt] + c

= -(1/2)*[x + log(sinx + cosx)] + c (By putting sinx + cosx = t)

So ∫{sinx/(sinx+cosx)}dx = -(1/2)*[x + log(sinx + cosx)] + c