# Integration by parts (trig integral)

Evaluate the following integral:

$$\int \cos(x)\cos(5x)\, dx$$

In the answers it says to solve by integration of parts twice with a consistent $u$ and $\frac{dv}{dx}$ to get $\frac{5}{24}\cos(x)\sin(5x) −\frac{1}{24}\sin(x)\cos(5x)+ C$.

I've tried using either value for $u$ or $\frac{dv}{dx}$ and can't seem to end up with the same answer. I keep ending up with the $(5x)$ still within $\cos$ or $\sin$ in the integral.

• Alternatively, you can do $$\int\Big(\frac{e^{ix}+e^{-ix}}{2}\Big)\Big(\frac{e^{5ix}+e^{-5ix}}{2}\Big)\,dx$$ May 22 at 15:17

$$I=\int \cos x \left(\frac{\cos (5x)\, d(5x)}{5}\right)$$

$$=(\cos x)\left(\frac{\sin(5x)}{5}\right)-\int (-\sin x)\left(\frac{\sin(5x)}{5}\right)\, dx$$

$$=(\cos x)\left(\frac{\sin(5x)}{5}\right)+\int (\sin x)\left(\frac{\sin(5x)\, d(5x)}{25}\right)$$

$$=(\cos x)\left(\frac{\sin(5x)}{5}\right)+\left((\sin x)\left(\frac{-\cos (5x)}{25}\right)-\int (\cos x)\left(\frac{-\cos (5x)}{25}\right)\, dx\right)$$

$$=(\cos x)\left(\frac{\sin(5x)}{5}\right)+(\sin x)\left(\frac{-\cos (5x)}{25}\right)+\frac{I}{25}$$

Solve for $I$.

$$I=\frac{25}{24}\left((\cos x)\left(\frac{\sin(5x)}{5}\right)+(\sin x)\left(\frac{-\cos (5x)}{25}\right)\right)+C$$

$$=\frac{5}{24}\cos x\sin(5x)-\frac{1}{24}\sin x\cos (5x)+C$$

Without using integration by parts: use the identity $$\cos x\cos y=\frac{1}{2}(\cos(x+y)+\cos(x-y))$$
$$\int \cos x\cos (5x)\,dx =\frac{1}{2}\int (\cos(6x)+\cos (4x))\, dx$$
$$=\frac{1}{2}\left(\frac{1}{6}\int \cos (6x)\, d(6x)+\frac{1}{4}\int \cos(4x)\, d(4x)\right)$$
$$=\frac{1}{12}\sin(6x)+\frac{1}{8}\sin(4x)+C$$