# 1/(sinx*(sinx+cosx)) integrate the following integral

(dy)/(dx) = 1/(sin(x)(sin(x)+cos(x)))

`y = int1/(sin(x)(sin(x)+cos(x)))dx`

Divide bothe numerator and denominator by sin(x), then,

`y = int(1/sin(x)/(sin(x)(sin(x)+cos(x))/sin(x))dx`

You get,

`y = int(cosec^2(x))/(1+cot(x))dx`

We know, d(cot(x) = -cosec^2(x)

Then using that in our integral, we can change the operator,

`y = -int1/(1+cot(x))d(cot(x))`

Now we can simply integrate this wrt cot(x),

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(dy)/(dx) = 1/(sin(x)(sin(x)+cos(x)))

`y = int1/(sin(x)(sin(x)+cos(x)))dx`

Divide bothe numerator and denominator by sin(x), then,

`y = int(1/sin(x)/(sin(x)(sin(x)+cos(x))/sin(x))dx`

You get,

`y = int(cosec^2(x))/(1+cot(x))dx`

We know, d(cot(x) = -cosec^2(x)

Then using that in our integral, we can change the operator,

`y = -int1/(1+cot(x))d(cot(x))`

Now we can simply integrate this wrt cot(x),

`y = -ln(1+cot(x))+c`

`y = -ln((sin(x)+cos(x))/sin(x))+c`

`y = ln((sin(x))/(sin(x)+cos(x)))+c`

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