Evaluate $\int\tan^2(x)\sec(x)\ dx$

Question

What would be the integration of $\displaystyle\tan^2(x)\sec(x)\ dx$? I have solved this by using reduction formula as $\displaystyle(\sec^2(x)-1)\sec(x)=\sec^3(x)-\sec(x)$ and then i apply reduction formula. but i want to know is there any other way to solve this rather than reduction formula??

Answer 1

Hint:

Take advantage of: $\frac{d}{dx}(\cos^{-2}{x}) = -2 \cos^{-3}{x} \, \sin{x}$ and write your integral as follows:

$$I = -\frac{1}{2} \int \sin{x} \, (-2 \, \sin{x} \cos^{-3} {x}) \, dx ,$$

apply now the chain rule to obtain:

$$I = -\frac{1}{2} \left( \frac{\sin{x}}{\cos^2{x}} - \int \frac{\cos{x}}{\cos^2 x} \, dx \right) = -\frac{1}{2} \left( \frac{\sin{x}}{\cos^2{x}} - \int \sec{x} \, dx \right) . $$ Recall now that:

$$\int \sec{x} \, dx = \ln{(\tan{x} + \sec{x})},$$

and substitute back in $I$.

Cheers!