Answer:
[tex]\tan\theta + \sec\theta + C[/tex]
Step-by-step explanation:
This cool problem uses an old trick: multiplying by a cleverly chosen expression for 1 (a fraction with the same numerator and denominator).
[tex]\int \frac{\sec\theta}{\sec\theta-\tan\theta} \cdot \frac{\sec\theta+\tan\theta}{\sec\theta+\tan\theta}\, d\theta\\\\=\int\frac{\sec\theta(\sec\theta+\tan\theta)}{\sec^2\theta-\tan^2\theta}[/tex]
That denominator looks kind of familiar. Remember one of the so-called Pythagorean identities?
[tex]1+\tan^2\theta=\sec^2\theta\\\\\sec^2\theta - \tan^2\theta =1[/tex] The denominator of the integrand is just 1 !!!
The integral is now
[tex]\int \sec\theta(\sec\theta+\tan\theta) \, d\theta = \int(\sec^2\theta+\sec\theta\tan\theta) \, d\theta\\=\int \sec^2\theta \, d\theta + \int \sec\theta\tan\theta \, d\theta\\=\tan\theta + \sec\theta + C[/tex]
That little trick is good to know. You may have used it before to rationalize a denominator. Example:
[tex]\frac{1}{\sqrt{7}-3}\cdot\frac{\sqrt{7}+3}{\sqrt{7}+3}=\frac{\sqrt{7}+3}{7-9}=\frac{\sqrt{7}+3}{-2}[/tex]
