Q.Choose the correct option: $\int \frac{dx}{\cos x\sqrt{\sin^2 x-\cos^2 x}} =$
(A) $\log(\tan x-\sqrt{\tan^2 x-1})+c$ (B) $\sin^{-1}(\tan x)+c$ (C) $1+\sin^{-1}(\cot x)+c$ (D) $\log(\tan x+\sqrt{\tan^2 x-1})+c$

Integrate by parts with $u=\sqrt{x^2+a^2}$, $dv=dx$, then use $x^2=(x^2+a^2)-a^2$.

Let $I=\displaystyle\int\sqrt{x^2+a^2}\,dx$. Integrating by parts with $u=\sqrt{x^2+a^2}$, $dv=dx$ (so $v=x$):

$I = x\sqrt{x^2+a^2} - \int x\cdot\dfrac{x}{\sqrt{x^2+a^2}}\,dx = x\sqrt{x^2+a^2}-\int\dfrac{x^2}{\sqrt{x^2+a^2}}\,dx$

Write $\dfrac{x^2}{\sqrt{x^2+a^2}} = \dfrac{(x^2+a^2)-a^2}{\sqrt{x^2+a^2}} = \sqrt{x^2+a^2}-\dfrac{a^2}{\sqrt{x^2+a^2}}$:

$I = x\sqrt{x^2+a^2} - \int\sqrt{x^2+a^2}\,dx + a^2\int\dfrac{dx}{\sqrt{x^2+a^2}}$

…

Integrate by parts with $u=\sqrt{a^2-x^2}$, $dv=dx$; the resulting integral reproduces $I$ itself, giving a solvable equation for $I$.

Let $I=\displaystyle\int\sqrt{a^2-x^2}\,dx$. Integrate by parts, taking $u=\sqrt{a^2-x^2}$ and $dv=dx$ (so $v=x$):

$I=x\sqrt{a^2-x^2}-\int x\cdot\frac{-x}{\sqrt{a^2-x^2}}\,dx=x\sqrt{a^2-x^2}+\int\frac{x^2}{\sqrt{a^2-x^2}}\,dx$

Write $x^2=-(a^2-x^2)+a^2$:

$\int\frac{x^2}{\sqrt{a^2-x^2}}dx=\int\frac{-(a^2-x^2)+a^2}{\sqrt{a^2-x^2}}dx=-\int\sqrt{a^2-x^2}\,dx+a^2\int\frac{dx}{\sqrt{a^2-x^2}}$

$=-I+a^2\sin^{-1}\left(\frac xa\right)$

So: …