Question

\[\int\frac{1}{p + q \tan x} \text{ dx  }\]

Answer 1

 

\[\text{ Let I }= \int\frac{dx}{p + q \tan x}\]
\[ = \int\frac{1}{p + \frac{q \sin x}{\cos x}}dx\]
\[ = \int\frac{\cos x}{q \sin x + p \cos x}dx\]
\[\text{ Let cos x} = A \left(\text{  q  sin x + p  cos x} \right) + B \left( q \cos x - p \sin x \right)\]
\[ \Rightarrow \cos x = \left( Ap + Bq \right) \cos x + \left( Aq - Bp \right) \sin x\]

Comparing coefficients of like terms

\[Ap + Bq = 1 . . . \left( 1 \right)\]
\[Aq - Bp = 0 . . . \left( 2 \right)\]

\[\Rightarrow A p^2 + Bpq = p\]
\[ \Rightarrow A q^2 - Bpq = 0\]
\[ \Rightarrow A = \frac{p}{p^2 q^2}\]

Putting value of A in eq (1)

\[\frac{p^2}{p^2 + q^2} + Bq = 1\]
\[ \Rightarrow Bq = 1 - \frac{p^2}{p^2 + q^2}\]
\[ \Rightarrow Bq = \frac{p^2 + q^2 - p^2}{p^2 + q^2}\]
\[ \Rightarrow B = \frac{q}{p^2 + q^2}\]
\[ \therefore I = \int\left[ \frac{p}{p^2 + q^2} \times \frac{\left( q \sin x + p \cos x \right)}{\left( q \sin x + p \cos x \right)} + \frac{q}{p^2 + q^2} \times \frac{\left( q \cos x - p \sin x \right)}{\left( q \sin x + p \cos x \right)} \right]dx\]
\[ = \frac{p}{p^2 + q^2}\int dx + \frac{q}{p^2 + q^2}\int\left( \frac{q \cos x - p \sin x}{q \sin x + p \cos x} \right)dx\]
\[\text{ Putting  q sin x + p  cos x = t}\]
\[ \Rightarrow \left( q \cos x - p \sin x \right) dx = dt\]
\[ \therefore I = \frac{p}{p^2 + q^2}\int\ dx + \frac{q}{p^2 + q^2}\int\frac{1}{t}dt\]
\[ = \frac{p}{p^2 + q^2} x + \frac{q}{p^2 + q^2} \text{ ln } \left| q \sin x + p \cos x \right| + C\]