# Find the integral of the function: $\int sin(3x). cos(4x) dx$

 Solution: $\frac{1}{2}. cosx - \frac{1}{14}. cos\;7x + k$

Toolbox.

• $\int sinx = - cos x$
• $sin(-x) = - sin(x)$
• Anytime you see a problem that is a product of a sin and cos function, use the following trignometric identity: $sin(\alpha).cos(\beta) = \frac{1}{2}. ((sin(\alpha-\beta)+sin(\alpha+\beta))$

Step-1
Let us use the information from the Toolbox above. Substitute $\alpha = 3x \, and \, \beta = 4x$.
Therefore the identity expands to $\frac{1}{2}. (sin(3x-4x) + sin(3x+4x))$, which is equal to $\frac{1}{2}.(sin(-x) + sin(7x))$
Remember that since $sin(-x) = - sin(x)$, we can rewrite this as $\frac{1}{2}. (sin(7x) - sin(x))$

Step-2
Hint: Whenever you see an equation with multiple terms, integrate the function term by term and factor out any constants.
i.e., Let $u = 7x$. Then $du = 7.dx$ and therefore $dx = \frac{1}{7}. du$
Then we can rewrite the function as $\frac{1}{2}. \frac{1}{7}.sin(u).du - \frac{1}{2}.sin(x).dx$

Step-3
Hint: Remember: $\int sinx = - cosx$.
Therefore, $\int (\frac{1}{2}. \frac{1}{7}.sin(u).du) =- \frac{1}{14}. cos(u)$
Therefore, $\int (- \frac{1}{2}.sin(x).dx) = + \frac{1}{2}.cos(x)$

Step-4
Evaluating the entire term above, we get $-\frac{1}{14}.cos(7x) + \frac{1}{2}.cos(x) + k (constant)$
Rewriting, we get the solution as $\frac{1}{2}. cosx - \frac{1}{14}. cos\;7x + k$

edited Dec 6, 2012