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

How would Integrate the following.

$\int \cos^3(x)\cos(2x)$

I did

$\int \cos^3(x)(1-2\sin^2(x))$

$2\int \cos^3(x)-\cos^3x\sin^2x$

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

$\int \cos^3(x)cos(2x) dx$

$\int \cos^3(x)(1-2\sin^2(x)) dx$

$\int \cos^2(x) \cdot \cos(x) (1-2\sin^2(x)) dx$

$\int (1-\sin^2x)\cdot \cos(x) \cdot(1-2\sin^2(x)) dx$

$\int ( \cos(x)- 3 \sin^2x \cdot \cos(x) + 2 sin^4x \cdot \cos(x) ) dx$

now using inverse chain rule which is $$\int f^n(x) \cdot f'(x) dx= f^{n+1}(x) +C$$

$\sin(x)- \sin^3x +2/5 \sin^5x + C$

• 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