Evaluating the Integral of sec(θ)/(sec(θ) - tan(θ))

This article focuses on the process of evaluating the integral:

The Problem and Initial Simplifications

To evaluate the integral:

I int frac{sec theta}{sec theta - tan theta} dtheta

The key to simplifying the integrand lies in recognizing the trigonometric identities:

sec θ 1/cos θ

tan θ sin θ/cos θ

By substituting these identities, we can rewrite the integrand:

sec theta - tan theta frac{1}{cos theta} - frac{sin theta}{cos theta} frac{1 - sin theta}{cos theta}

Further Simplification

Re-substituting the simplified integrand into the integral, we have:

I int frac{sec theta}{frac{1 - sin theta}{cos theta}} dtheta int frac{sec theta cos theta}{1 - sin theta} dtheta int frac{1}{1 - sin theta} dtheta

Multiplying by the Conjugate

To simplify further, we use the technique of multiplying the numerator and denominator by the conjugate of the denominator:

frac{1}{1 - sin theta} cdot frac{1 sin theta}{1 sin theta} frac{1 sin theta}{1 - sin^2 theta} frac{1 sin theta}{cos^2 theta}

Thus, the integral becomes:

I int frac{1 sin theta}{cos^2 theta} dtheta int sec^2 θ dθ - int frac{sin θ}{cos^2 θ} dθ

Breaking Down the Integral

We then break down the integral into two simpler parts:

The first part is the integral of sec2θ

The second part involves the substitution for the product of tanθ and secθ

The Integral of sec2θ

The integral of sec2θ is a standard result:

int sec^2 θ dθ tan θ C_1

The Second Integral: tanθsecθ

For the second integral, we use the substitution:

u sec θ, du sec θ tan θ dθ

Thus:

int tan θ sec θ dθ int du u C_2 sec θ C_2

Combining Results

Combining both parts, we have:

I tan θ sec θ C, where C C_1 C_2

Therefore, the final result for the integral is:

int frac{sec θ}{sec θ - tan θ} dθ tan θ sec θ C

Conclusion

The integral evaluation of sec(θ)/(sec(θ) - tan(θ)) involves a series of simplifications, clever manipulations of trigonometric identities, and the use of standard integral results.