Integrating $\sin 2x \cos 2x$
I am having trouble solving the following:
$$\int \sin2x\cos2x\,dx$$
I understand an easy way of going about this is using the trig identity $\sin2x = 2\sin x\cos x$, so I thought that it would be:
$$\int \frac{\sin 2x}{2} dx$$
which would then give me $-\cos 2x / 4$.
However, in the maths book I am going from, they do it this way:
$$\int \sin 4x dx$$
which then gives an answer of $-\cos 4x / 4 + c$, which differs from my answer and answers obtained from online integral calculators.
Any help would be appreciated, thanks.
$\endgroup$ 26 Answers
$\begingroup$First, the formula is
$$\sin{4x}=2\sin{2x}\cos{2x}$$
Second, your integral becomes
$$\int \frac{\sin{4x}}{2} dx$$
Then the answer is
$$-\frac{\cos{4x}}{8}+C$$
$\endgroup$ 0 $\begingroup$Hint
To compute the integral $$I=\int \sin(2x)\cos(2x)~dx$$, there is two possible ways.
- the first one : use $\sin(2a)=2\sin(a)\cos(a)$ and so, $$I=\frac 12 \int \sin(4x)~dx=-\frac 12\times\frac 14 \cos(4x)+C=-\frac 18 \cos(4x)+C$$
- the second one : change variable $u=\sin(2x)$, $du=2\cos(2x)~dx$ and so $$I=\frac 12\int u ~du=\frac{u^2}4+C=\frac 14\sin^2(2x)+C$$ Now use $2\sin^2(a)=1-\cos(2a)$ and you will arrive to the same result, the constant term being absorbed in $C$
$$\int \sin2x\cos2x\,dx$$ $$(\sin2x=t,\cos2xdx=dt/2)$$ $$\int \frac{t}{2}dt=\frac{t^2}{4}+C$$
$\endgroup$ $\begingroup$Rearranging the double angle identity gives $$\sin u \cos u = \frac{1}{2} \sin 2u,$$ so substituting $u = 2x$ gives $$\int \sin 2x \cos 2x \,dx = \frac{1}{2} \int \sin 4x \,dx = -\frac{1}{8} \cos 4x + C.$$
The different-looking forms you've found for the antiderivatives might be a consequence of the following: We can alternatively substitute $v = \sin 2x$, $dv = 2 \cos 2x \,dx$, so the antiderivative is $$\frac{1}{2} \int v \,dv = \frac{1}{4} v^2 + C = \frac{1}{4} \sin^2 2x + C.$$ But, using a double-angle identity shows that this agrees with the answer produced above.
$\endgroup$ $\begingroup$Imagine the 2x's became just x. You could use the substitution u = sin x or u = cos x or change the integrand to $\frac{\sin 2x}{2} dx$.
So let's substitute t = 2x. Then we have $\frac{1}{2} \int \sin t \cos t dt$.
$\endgroup$ $\begingroup$In general, to compute the integrals of the form $$\int \sin ax\cos bx \,dx$$ convert the product to sum using the identity $$\sin ax\cos bx={1\over 2}\{sin(a+b)x+\sin(a-b)x\}$$ Integrating RHS of the latter is easy.
$\endgroup$
