By using the substitution $u = \sec x$, or otherwise, find \[ \int \sec^3 x \tan x \, dx \] - HSC - SSCE Mathematics Extension 2 - Question 1 - 2002 - Paper 1

First, we complete the square for the expression in the denominator: [ x^2 + 2x + 2 = (x + 1)^2 + 1. ]

This leads to: [ \int \frac{dx}{(x + 1)^2 + 1} = \tan^{-1}(x + 1) + C. ]

To solve this integral, we use partial fraction decomposition: [ \frac{x}{(x + 3)(x - 1)} = \frac{A}{x + 3} + \frac{B}{x - 1}. ]

By solving for A and B, we find: [ A = \frac{1}{4}, B = -\frac{1}{4}. ]

The integral then can be solved as: [ \int \left( \frac{1/4}{x + 3} - \frac{1/4}{x - 1} \right) dx = \frac{1}{4} \ln |x + 3| - \frac{1}{4} \ln |x - 1| + C = \frac{1}{4} \ln \left| \frac{x + 3}{x - 1} \right| + C. ]

Letting $u = e^x$ and $dv = \cos x \, dx$, we find: [ du = e^x , dx, \quad v = \sin x. ]

Thus, applying integration by parts twice: [ \int e^x \cos x , dx = e^x \sin x - \int e^x \sin x , dx. ]

Continuing this process leads us to: [ I = \int e^x \cos x , dx, J = \int e^x \sin x , dx]

After integrating and substituting back, we evaluate from $0$ to $\frac{\pi}{2}$. The final answer simplifies to: [ I + J = e^{\frac{\pi}{2}} \cdots + \cdots. ]

Using the Weierstrass substitution where: [ \cos \theta = \frac{1 - t^2}{1 + t^2}, \quad d\theta = \frac{2 , dt}{1 + t^2}, ]

We can rewrite the integral as: [ \int \frac{2 , dt}{(2 + \frac{1 - t^2}{1 + t^2})(1 + t^2)}]

This leads to a new expression which can be simplified and evaluated. Ultimately we find: [ \text{Final evaluation from 0 to } \infty = \cdots + C. ]