Q.Find the integral of the function $\frac{1}{\cos(x-a)\cos(x-b)}$

Introduce the constant $\sin(a-b)$ so the integrand splits into two tangents.

Since $(x-b)-(x-a)=a-b$, we have $\sin(a-b)=\sin\big((x-b)-(x-a)\big)=\sin(x-b)\cos(x-a)-\cos(x-b)\sin(x-a)$. Divide this by $\sin(a-b)\cos(x-a)\cos(x-b)$:

Using $\int\tan u\,du=-\log\lvert\cos u\rvert$, …

The trick

There is no obvious substitution for a product of two shifted cosines. The standard move is to manufacture a sine in the numerator using the constant angle $a-b$. Notice $(x-b)-(x-a)=a-b$, so $\sin(a-b)$ is a constant we can insert for free.

Build the numerator

Expand

Divide both sides by $\sin(a-b)\,\cos(x-a)\cos(x-b)$:

Integrate

With $\int\tan u\,du=-\log\lvert\cos u\rvert$, …

Method: Multiply by $\dfrac{\sin(a-b)}{\sin(a-b)}$ to split $\dfrac{1}{\cos(x-a)\cos(x-b)}$

For a reciprocal of two cosines with different phase shifts, manufacture the constant $\sin(a-b)$ so the numerator becomes a difference that splits into two tangents.

Steps

Step 1: Insert the constant cleverly.

Multiply and divide by $\sin(a-b)$, and write $a-b=(x-b)-(x-a)$:

Step 2: Expand the sine of a difference and split.

Using $\sin(P-Q)=\sin P\cos Q-\cos P\sin Q$ with $P=x-b$, $Q=x-a$, the fraction separates into

$\frac{1}{\sin(a-b)}\big[\tan(x-b)-\tan(x-a)\big]$ …

Common Mistakes

Mistake 1: Not knowing to manufacture $\sin(a-b)$ in the numerator.

Why it's wrong: without introducing $\sin(a-b)=\sin[(x-b)-(x-a)]$, the product of two cosines cannot be split. Correct approach: multiply and divide by $\sin(a-b)$ and expand the sine of the angle difference.

Mistake 2: Sign slip when integrating $\tan$. …