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.

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

1 Answer 1


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

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

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

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

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

Solve for $I$.

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

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

This agrees with your book's answer.

Without using integration by parts: use the identity $$\cos x\cos y=\frac{1}{2}(\cos(x+y)+\cos(x-y))$$

$$\int \cos x\cos (5x)\,dx =\frac{1}{2}\int (\cos(6x)+\cos (4x))\, dx$$

$$=\frac{1}{2}\left(\frac{1}{6}\int \cos (6x)\, d(6x)+\frac{1}{4}\int \cos(4x)\, d(4x)\right)$$


  • $\begingroup$ Which is nicer that the answer from the book, for sure ! $\endgroup$ Oct 24, 2015 at 4:54
  • $\begingroup$ Alright thanks!, the identity way does seem a whole lot easier. $\endgroup$
    – James
    Oct 24, 2015 at 5:05
  • $\begingroup$ In general , if you have this type of problem where you have two trig functions multiplied together as the integrand, you should use the trig identity, this is a very standard trick. $\endgroup$
    – ccchanah
    Oct 24, 2015 at 5:09

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .