Q.Integrate the following function: $\sin (ax+b) \cos (ax+b)$

The key idea is to use the Product-to-Sum identity to rewrite the product as a single sine term.

Step 1: Recall the identity:

Here, $\theta = ax + b$.

Step 2: Rewrite the integrand:

Step 3: Integrate: …

Why This Approach Works

When you see a product of sine and cosine with the same argument (here both are $ax+b$), your first instinct might be to try substitution. But there's a cleaner path. The product $\sin \theta \cos \theta$ is actually half of $\sin 2\theta$ — that's a standard double-angle identity in reverse. This transforms the integral from a product into a simple sine function, which integrates to a cosine. No messy u-substitution needed, and the algebra stays minimal.

The identity we need is:

So $\sin \theta \cos \theta = \frac{1}{2} \sin 2\theta$. Here $\theta = ax+b$.

Step-by-Step Solution

1. Apply the identity.

Let $\theta = ax+b$. Then:

2. Set up the integral.

The integral becomes:

3. Integrate the sine function.

Recall that $\int \sin(kx + c) \, dx = -\frac{1}{k} \cos(kx + c) + C$. Here $k = 2a$ and $c = 2b$. So:

4. Multiply by the constant factor.

$\frac{1}{2} \left( -\frac{1}{2a} \cos(2ax + 2b) + C_1 \right) = -\frac{1}{4a} \cos(2ax + 2b) + C$ …

Method: Collapse a trig product with a double-angle identity

Use this when the integrand is a product of a sine and cosine of the same angle, e.g. $\sin(ax+b)\cos(ax+b)$ — turn the product into a single sine before integrating.

Steps

Step 1: Apply $\sin\theta\cos\theta=\tfrac12\sin 2\theta$.

With $\theta=ax+b$,

Step 2: Integrate the single sine with the linear-argument rule.

$\int\sin(k\,x+c)\,dx=-\frac{1}{k}\cos(kx+c)+C,$ …

Common Mistakes

Mistake 1: Integrating the product $\sin\cos$ term by term.

Why it's wrong: there is no rule to integrate $\sin(ax+b)\cos(ax+b)$ as a product. Correct approach: use $\sin\theta\cos\theta=\tfrac12\sin2\theta$ to convert to a single sine first.

Mistake 2: Forgetting the $\dfrac{1}{2a}$ from the linear argument.

Why it's wrong: $\int\sin(2ax+2b)\,dx=-\dfrac{1}{2a}\cos(2ax+2b)$; the coefficient of $x$ is $2a$, not $1$. Correct approach: divide by the coefficient of $x$ in the angle. …