Q.Integrate the following function: $\sin 3x \cos 4x$

Turn the product into a sum with $\sin A\cos B=\frac12[\sin(A+B)+\sin(A-B)]$.

With $A=3x,\ B=4x$:

Integrate term by term (using $\int\sin kx\,dx=-\frac{1}{k}\cos kx$):

Why convert to a sum

Products of sines and cosines are hard to integrate directly, but sums are trivial. The identity $\sin A\cos B=\frac12[\sin(A+B)+\sin(A-B)]$ does the conversion.

Apply the identity

Take $A=3x$, $B=4x$:

Since $\sin(-x)=-\sin x$,

Integrate

Use $\int\sin kx\,dx=-\frac{1}{k}\cos kx$:

Method: Product-to-sum identity for $\sin(ax)\cos(bx)$

A product of a sine and a cosine of different angles cannot be integrated as-is; convert it into a sum of sines, which integrate term by term.

Steps

Step 1: Apply the correct product-to-sum identity.

Match $A$ and $B$ to the two angles in the integrand.

Step 2: Simplify signed angles using parity.

A negative angle inside sine flips sign: $\sin(-x)=-\sin x$. (For cosine, $\cos(-x)=\cos x$.) Simplify before integrating.

Step 3: Integrate each sine term.

Use $\displaystyle\int\sin(kx)\,dx=-\frac1k\cos(kx)+C$, keeping the $\dfrac1k$ factor for every term.

Remember which identity to reach for: a $\sin\cos$ or $\cos\sin$ product yields sines, while $\sin\sin$ or $\cos\cos$ yields cosines.

Common Mistakes

Mistake 1: Trying to integrate $\sin 3x\cos 4x$ as a single product.

Why it's wrong: there is no $u$ whose derivative appears because the two angles differ, so direct substitution fails. Correct approach: use $\sin A\cos B=\tfrac12[\sin(A+B)+\sin(A-B)]$ to split it into $\tfrac12[\sin 7x+\sin(-x)]$.

Mistake 2: Leaving $\sin(-x)$ without simplifying its sign.

Why it's wrong: $\sin(-x)=-\sin x$, so overlooking the odd symmetry gives a wrong sign on that term. Correct approach: simplify to $\tfrac12[\sin 7x-\sin x]$ before integrating, yielding $-\tfrac{1}{14}\cos 7x+\tfrac12\cos x+C$.