How do you find the integral of #sec^5 x dx#?

To find the integral of ( \sec^5(x) , dx ), you can use integration by parts method. The integral can be expressed as:

[ \int \sec^5(x) , dx = \int \sec(x) \cdot \sec^4(x) , dx ]

Let ( u = \sec(x) ) and ( dv = \sec^4(x) , dx ).

Then, ( du = \sec(x) \tan(x) , dx ) and ( v = \frac{1}{3} \tan(x) \sec^3(x) ).

Applying the integration by parts formula ( \int u , dv = uv - \int v , du ), we get:

[ \int \sec^5(x) , dx = \frac{1}{3} \sec(x) \tan(x) \sec^3(x) - \frac{1}{3} \int \tan(x) \sec^3(x) \sec(x) \tan(x) , dx ]

Simplify the integral:

[ \int \sec^5(x) , dx = \frac{1}{3} \sec(x) \tan(x) \sec^3(x) - \frac{1}{3} \int \sec^4(x) , dx ]

Now, the integral ( \int \sec^4(x) , dx ) can be computed using a power-reduction formula or another application of integration by parts.

After finding the integral, you can substitute back the original variable ( x ).