# Integrating $\int \cos^3(x)\cos(2x) \, dx$

How would Integrate the following? $$\int \cos^3(x)\cos(2x) \, dx.$$ I did $$\int \cos^3(x)(1-2\sin^2(x)) \, dx = 2\int \cos^3(x)-\cos^3x\sin^2x \, dx$$

But I find myself stuck....

• Don't leave out $dx$ in the integrals! When you make a change of variables, the $dx$ part is essential to get things right. – Mårten W Jan 23 '14 at 16:15

That's a good way to proceed. So our integral is $$\int \cos^3 x(1-2\sin^2 x)\,dx.$$

Rewrite as $$\int \cos x(1-\sin^2 x)(1-2\sin^2 x)\,dx$$ and let $u=\sin x$. We end up with $$\int (1-u^2)(1-2u^2)\,du.$$ Expand and integrate.

• Yes yes I will try this – Fernando Martinez Sep 18 '13 at 18:36
• But where would the cosx go do I just put it back in after expanding? – Fernando Martinez Sep 18 '13 at 18:37
• or if u=sin(x) the du=cos(x) – Fernando Martinez Sep 18 '13 at 18:41
• Expand. We get $1-3u^2+2u^4$. Integrate. We get $u-u^3+\frac{2}{5}u^5+C$. Now replace $u$ by $\sin x$. About the $\cos x$, that disappeared during the substitution. We had let $u=\sin x$. So $du=\cos x \,dx$. Thus $\int\cos x(1-\sin^2 x)(1-2\sin^2 x)\,dx=\int(1-u^2)(1-2u^2)\,du$. – André Nicolas Sep 18 '13 at 18:41
• I see thanks for your well explained and comprehensive answer.. – Fernando Martinez Sep 18 '13 at 18:41

HINT:

As $\cos3x=4\cos^3x-3\cos x$

$\displaystyle\cos^3x\cos2x=\frac{(\cos3x+3\cos x)\cos2x}4=\frac{2\cos3x\cos2x+3\cdot2\cos2x\cos x}8$

Use $2\cos A\cos B=\cos(A-B)+\cos(A+B)$ and $\displaystyle\int\cos mxdx=\frac{\sin mx}m+K$ where $K$ is an arbitrary constant

\begin{align} \int \cos^3(x) \cos(2x) dx &= \int \cos^3(x)(1-2\sin^2(x)) dx \\ &= \int \cos^2(x) \cdot \cos(x) (1-2\sin^2(x)) dx \\ &= \int (1-\sin^2x)\cdot \cos(x) \cdot(1-2\sin^2(x)) dx \\ &= \int ( \cos(x)- 3 \sin^2x \cdot \cos(x) + 2 \sin^4x \cdot \cos(x) ) dx. \end{align}

Now using inverse chain rule which is $$\int f^n(x) \cdot f'(x) dx= f^{n+1}(x) +C$$ then $$\int \cos^3(x) \cos(2x) dx = \sin(x)- \sin^3x + \frac{2}{5} \, \sin^5x + C$$

• Tip: Use a \ before 'sin' or 'cos' – The Chaz 2.0 Sep 18 '13 at 18:35
• ok thanks will edit – MRK Sep 18 '13 at 18:36