Evaluate: `int1/(p+qtanx)dx`

Explanation

`int1/(p+qtanx)dx=int1/(p+frac{qsinx}{qcosx})dx`

`=int cosx/(q sinx+p cosx)dx`

Let `cos x=A(q sinx+p cosx)+B(q cosx-p sin x)=>(Ap+Bq)cosx+(Aq-Bp)sinx`

so, `Ap+Bq=1 ---(1)`

`Aq-Bp=0 ---(2)`

With the help of eq,(1) and (2), we get, `A=frac{p}{p^2q^2}` and `B=frac{q}{p^2q^2}`

`I=int(frac{p}{p^2q^2} xx frac{qsinx_pcosx}{qsinx+pcosx})+frac{q}{p^2q^2} xx frac{q cosx-p sinx}{qsinx+pcosx}`

Putting, `q sinx+p cos x=t`
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