Integrals Involving Trig Functions — Question 6

Question 6 – Solution

Use the product-to-sum identity $$\sin(A)\cos(B)=\frac12\bigl[\sin(A+B)+\sin(A-B)\bigr].$$ With $A=5x$ and $B=2x$, $$\sin(5x)\cos(2x)=\frac12\bigl[\sin(7x)+\sin(3x)\bigr].$$ Therefore, $$\begin{align*} \int\sin(5x)\cos(2x)\,dx &=\frac12\int\bigl[\sin(7x)+\sin(3x)\bigr]\,dx\\ &=\frac12\left(-\frac17\cos(7x)-\frac13\cos(3x)\right)+C\\ &=-\frac1{14}\cos(7x)-\frac16\cos(3x)+C. \end{align*}$$ Hence, $$\boxed{\displaystyle \int\sin(5x)\cos(2x)\,dx =-\frac1{14}\cos(7x)-\frac16\cos(3x)+C}.$$ Differentiating gives $$\frac12\sin(7x)+\frac12\sin(3x) =\sin(5x)\cos(2x),$$ confirming the answer by the same product-to-sum identity.