Question

$\int \frac{secx\cdot dx}{b+atanx}$

Answer 1

$\int \frac{secxdx}{b+atanx}$

$=\int \frac{1/cosx}{(bcosx+asinx)/cosx}dx$

$=\int \frac{1}{bcosx+asinx}dx$

$=\int \frac{1}{\sqrt{a^2+b^2\{\frac{b}{\sqrt{a^2+b^2}}cosx+\frac{a}{\sqrt{a^2+b^2}}sinx\}}}dx$

$=\frac{1}{\sqrt{a^2+b^2}}\int \frac{1}{\{\frac{b}{\sqrt{a^2+b^2}}cosx+\frac{a}{\sqrt{a^2+b^2}}sinx\}}dx$

Let $\frac{b}{\sqrt{a^2+b^2}}=sinA$ and $\frac{a}{\sqrt{a^2+b^2}}=cosA$

We can see that $\frac{b^2}{{\left(\sqrt{a^2+b^2}\right)}^2}+\frac{a^2}{{\left(\sqrt{a^2+b^2}\right)}^2}=1={\sin }^{2}A{\cos }^{2}A$, So our choice is justified. Substituting we get

$=\frac{1}{\sqrt{a^2+b^2}}\int \frac{1}{sinAcosx+cosAsinx}dx$

$=\frac{1}{\sqrt{a^2+b^2}}\int \frac{1}{\sin (A+x)}dx$ $(\because \sin (A+B)=sinA.cosB+cosA.sinB)$

$=\frac{1}{\sqrt{a^2+b^2}}\int cosec(A+x)dx$

$=\frac{1}{\sqrt{a^2+b^2}}ln\left|\tan (A+x)\right|+c$

$=\frac{1}{\sqrt{a^2+b^2}}ln\left|\tan (x+{\sin }^{-1}\frac{b}{\sqrt{a^2+b^2}})\right|+c$