Evaluation of $\int \sec^3 (x)dx$

Question

How Can I evaluate $\displaystyle \int \sec^3 (x)dx$

(Without Using Weierstrass Substution or Integration by parts.)

$\bf{My\; Try::}$ Let $\displaystyle I = \int\sec^3(x)dx = \int \frac{1}{\cos^3(x)}dx = \int \frac{1}{\sin ^3\left(\frac{\pi}{2}-x\right)}dx$

Now Let $\displaystyle \left(\frac{\pi}{2}-x\right) = t\;,$ Then $\displaystyle dx = -dt$. So $\displaystyle I = -\int \frac{1}{\sin^3 t}dt = -\int\frac{1}{2\sin^3\left(\frac{t}{2}\right)\cdot \cos^3 \left(\frac{t}{2}\right)}dt$

Now How can I solve after that

Help me

Thanks

Answer 1

Hint:

Substitite

$$\sec (x)= \cosh (u) $$

$$\tan(x) = \sinh(u)$$

$$\sec^2 (x) \, \mathrm dx = \cosh (u) \,\mathrm du$$

$$\mathrm dx = \frac1{\cosh(u)} \, \mathrm du$$

Edit: Why was this downvoted?

Here is the entire process

$$\begin{align} \int \sec^3 x \, \mathrm dx &{}= \int \cosh^2 u\,\mathrm du \\ &= \frac{1}{2}\int ( \cosh 2u +1) \,\mathrm du \\ &= \frac{1}{2} \left( \frac{1}{2}\sinh2u + u\right) + C\\ &= \frac{1}{2} ( \sinh u \cosh u + u ) + C \\ &= \frac{1}{2} \sec x \tan x + \frac{1}{2} \ln|\sec x + \tan x| + C \end{align} $$

Answer 2

If you don't mind partial fractions, you can simply rewrite your integral as

$$\int\frac{\cos x}{\cos^4x}dx=\int\frac{\cos x}{(1-\sin^2x)^2}dx$$

then substitute $u=\sin x$.