Evaluvate the following intregals: 1 p+q dx

Question

Evaluvate the following intregals:
$\int\frac{1}{\text{p}+\text{q}\tan\text{x}}\ \text{dx}$

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Answer

Let $\text{I}=\int\frac{1}{\text{p}+\text{q}\tan\text{x}}\ \text{dx}$
$=\int\frac{1}{\text{P}+\frac{q\sin\text{x}}{\cos\text{x}}}\ \text{dx}$
$=\int\frac{\cos\text{x}}{q\sin\text{x}+\text{p}\cos\text{x}}\ \text{dx}$
Let $\cos\text{x}=\text{A}(q\sin\text{x}+\text{p}\cos\text{x})+\text{B}(q\cos\text{x}-\text{p}\sin\text{x})$
$\Rightarrow\cos\text{x}=(\text{Ap}+\text{Bq})\cos\text{x}+(\text{Aq}-\text{Bp})\sin\text{x}$
Compairing coefficient of like terms
$\text{Ap}+\text{Bq}=0\dots(1)$
$\text{Aq}+\text{Bp}=1\dots(2)$
Multipiying eq (1) by p and eq (2) by q and then adding
$\Rightarrow\text{Ap}^2+\text{Bpq}=\text{p}$
$\Rightarrow\text{Aq}^2+\text{Bpq}=0$
$\Rightarrow\text{A}=\frac{\text{p}}{\text{p}^2\text{q}^2}$
Putting value of A in eq (1)
$\frac{\text{p}^2}{\text{p}^2+\text{q}^2}+\text{Bq}=1$
$\Rightarrow\text{Bq}=1-\frac{\text{p}^2}{\text{p}^2+\text{q}^2}$
$\Rightarrow\text{Bq}=\frac{\text{p}^2+\text{q}^2-\text{p}^2}{\text{p}^2+\text{q}^2}$
$\Rightarrow\text{B}=\frac{\text{q}}{\text{p}^2+\text{q}^2}$
$\therefore\text{I}=\int\Big[\frac{\text{p}^2}{\text{p}^2+\text{q}^2}\times\frac{(\text{q}\sin\text{x}+\text{p}\cos\text{x})}{(q\sin\text{x}+\text{p}\cos\text{x})}+\frac{\text{q}}{\text{p}^2+\text{q}^2}\times\frac{(q\cos\text{x}-\text{p}\sin\text{x})}{(q\sin\text{x}+\text{p}\cos\text{x})}\Big]\text{dx}$
$=\frac{\text{q}}{\text{p}^2+\text{q}^2}\int\text{dx}+\frac{\text{p}^2}{\text{p}^2+\text{q}^2}\int\Big(\frac{q\cos\text{x}-\text{p}\sin\text{x}}{q\sin\text{x}+\text{p}\cos\text{x}}\Big)\text{dx}$
Putting $\text{q}\sin\text{x}+\text{p}\cos\text{x}=\text{t}$
$\Rightarrow(\text{q}\cos\text{x}-\text{p}\sin\text{x})\text{ dx}=\text{dt}$
$\therefore\text{I}=\frac{\text{p}}{\text{q}+\text{q}^2}\int\text{dx}+\frac{\text{q}}{\text{p}^2+\text{q}^2}\int\frac{1}{\text{t}}\text{dt}$
$=\frac{\text{p}}{\text{q}+\text{q}^2}\text{x}+\frac{\text{q}}{\text{p}^2+\text{q}^2}\ln|\text{q}\sin\text{x}+\text{p}\cos\text{x}|+\text{C}$