Q.Integrate the function $\frac{1}{\cos(x+a)\cos(x+b)}$

The key idea is to rewrite the product of cosines using the identity for the difference of sines, then integrate via U Substitution.

We start with the identity:

Using $\sin(A-B) = \sin A \cos B - \cos A \sin B$, the numerator becomes $\sin(x+a)\cos(x+b) - \cos(x+a)\sin(x+b)$. Dividing by the denominator gives:

Now integrate term by term: …

Why This Approach Works

When you see a product of cosines in the denominator, your first instinct might be to try trigonometric identities. The product $\cos(x+a)\cos(x+b)$ looks like it could be part of a sum-to-product formula, but that leads to messy expressions. Instead, think about differentiation: the derivative of $\tan\theta$ is $\sec^2\theta$, and $\sec\theta = 1/\cos\theta$. If we could somehow express $1/(\cos A \cos B)$ as a difference of two tangent derivatives, the integral becomes trivial.

The trick lies in the identity:

If we set $A = x+a$ and $B = x+b$, then $A-B = a-b$, a constant. This means:

Now divide both sides by $\cos(x+a)\cos(x+b)$:

This is the breakthrough: the constant $\sin(a-b)$ factors out, leaving a simple difference of tangents. The integral then becomes straightforward.

Step-by-Step Solution

1. Set up the integral and apply the key identity

We want:

Multiply numerator and denominator by $\sin(a-b)$:

Using the identity derived above:

So:

2. Integrate the difference of tangents

Each tangent integrates to a logarithm:

$\int \tan(x+a) \, dx = -\log|\cos(x+a)| + C_1$ …

Method: Introduce $\sin(a-b)$ to split $\dfrac{1}{\cos(x+a)\cos(x+b)}$ into tangents

Use this for a product of two cosines (or sines) with constant-shifted arguments in the denominator: the constant angle-difference lets you write the integrand as a difference of tangents.

Steps

Step 1: Multiply and divide by $\sin(a-b)$.

Since $a-b=(x+a)-(x+b)$ is constant, $\sin(a-b)$ is just a number you may insert freely.

Step 2: Expand $\sin\bigl((x+a)-(x+b)\bigr)$ and divide by $\cos(x+a)\cos(x+b)$.

This yields …

Common Mistakes

Mistake 1: Treating $\sin(a-b)$ as depending on $x$.

Why it's wrong: $a-b$ is constant, so $\sin(a-b)$ is a number that factors out; treating it as variable breaks the method. Correct approach: pull $\tfrac{1}{\sin(a-b)}$ outside the integral.

Mistake 2: Sign error integrating tangent.

Why it's wrong: $\int\tan(x+c)\,dx=-\log|\cos(x+c)|$; dropping the minus inverts the final log ratio. Correct approach: use the negative log-cosine form. …