How to find the right answer for Integral of $\sin(2x)\cos(2x)$

Question

This is what I did when I solved the $$\int(\sin(2x)\cos(2x))dx$$

First I used integration and made $$u=2x$$ Then I found the derivative of $2x$ and determined its value when it is equal to $dx$, $$\frac{du}{dx}=2$$$$dx=\frac{du}{2}$$

Substituting these values into the integral we get $$\int[\sin(u)\cos(u)]\frac{du}{2}$$

Then I turned sine and cosine into their anti-derivatives. From my math class I learned that the anti-derivative of $\sin(x)$ is $-\cos x$ and the antiderivative of $\cos(x)$ is $\sin(x)$.

Thus my answer is

$$\frac{1}{2}[-\cos(2x)\sin(2x)+c]$$ However the answer is supposed to be $$\frac{\sin^2(2x)}{4}+C$$ Could someone tell me why?

Answer 1

You can't turn them into their antiderivatives when two functions are in multiplication. You have to simplify the integral until you know the antiderivative of the whole function inside the integral.

$$\int[\sin(u)\cos(u)]\frac{du}{2}$$ derivative of $\sin(u)$ is $\cos(u)$. We can solve the integral using the substitution
$$t=\sin(u)$$ $$dt=\cos(u)du$$

$$\int[\sin(u)\cos(u)]\frac{du}{2}=\int\frac{t}{2}dt=\frac{t^2}{4}=\frac{\sin^2(u)}{4}+C=\frac{\sin^2(2x)}{4}+C$$

Answer 2

Remember the antiderivative is the function whose derivative is the integrand. So a great way to check your answer is to take the derivative of it and see if you get the original question. If we try taking the derivative of your answer,

$\begin{align} \frac{d}{dx}\left[\frac{1}{2}(-\cos (2x)\sin(2x))\right]&=-\frac{1}{2}\left[\left(\frac{d}{dx}\cos(2x)\right)\sin(2x)+\cos(2x)\left(\frac{d}{dx}\sin(2x)\right)\right] \quad \text{Product rule!} \\ &=-\frac{1}{2}[-2\sin(2x)\sin(2x)+2\cos(2x)\cos(2x)]\\ &=-\frac{1}{2}[2\cos(4x)] \quad \text{Double angle identity $\cos(2\theta)=\cos^2\theta-\sin^2\theta$}\\ &=-\cos(4x) \end{align}$

This is not the same as the original integrand, $\sin(2x)\cos(2x)$. (Incidentally this can be simplified to $\frac{1}{2}\sin(4x)$ using the double angle identity of sine, and your answer is still not the same as the original integrand). The product rule is the crucial reason why the antiderivative of the product of the two functions is not the product of the antiderivatives. Like another answer pointed out, at the step $\displaystyle\int[\sin(u)\cos(u)]\frac{du}{2}$, instead of taking the antiderivative of $\sin(u)$ and $\cos(u)$, make a substitution $v=\sin(u)$ and $dv=\cos(u)\,du$. Then we get:

$$\int[\sin(u)\cos(u)]\frac{du}{2}=\int\frac{v}{2}dv=\frac{v^2}{4}+C $$.

Substituting back in $v=\sin(u)$ and $u=2x$, the final answer is $\displaystyle\frac{\sin^2 (2x)}{4}+C$.

Notice what happens when you take the derivative:

$\begin{align} \frac{d}{dx}\left[\frac{\sin^2 (2x)}{4}\right]&=\frac{1}{4}\cdot 2\sin(2x)\left(\frac{d}{dx}\sin(2x)\right) \quad \text{Chain Rule!}\\ &=\frac{1}{2}\sin(2x)\cos(2x)\left(\frac{d}{dx}(2x)\right) \quad \text{Chain rule again!} \\ &=\sin(2x)\cos(2x) \end{align}$

We got the same thing as our original integrand! Notice I had to use the chain rule twice, and in the original integral, you had to use substitution twice. Substitution is basically like doing the chain rule backwards, so that's why doing two substitutions got us to our antiderivative in the same way doing two iterations of the chain rule got us our integrand.

Answer 3

Here's a very simple substitution.

Use the fact that $\cos(2x)=1-2\sin^2x$. Notice that if you differentiate $\sin(2x)$ you will have another $\cos(2x)$ term, so you simply use the previous identity again.

Choose $u=\sin(2x)$. Then $\cfrac{du}{dx}=2\cos(2x)=2(1-2\sin^2x)$, then

$\begin{align*}\ \int\sin(2x)\cos(2x)dx &= \int \sin(2x)(1-2\sin^2x)dx \\ &= \int \cfrac u2 du \\ &=\cfrac{u^2}{4}+C, \hspace{1cm} u=\sin(2x) \\ &=\cfrac{\sin^2(2x)}{4}+C \end{align*}$

Now: the reason your attempt did not work is because $$\int f(x)g(x)dx \ne \int f(x)dx \cdot \int g(x)dx$$. Look at the product rule of the functions $f(x), g(x)$:

$\begin{align*} \cfrac{d}{dx}[f(x)g(x)] & = f'(x)g(x) + g'(x)f(x) \\ f(x)g(x) &= \int f'(x)g(x)dx + \int g'(x)f(x)dx \end{align*}$

Alternatively you may use the fact $\sin(\alpha + \beta) = \sin\alpha \cos\beta+\cos\alpha\sin\beta$. Letting $\alpha=\beta = 2x$ then $\sin(2x) \cos(2x) = \cfrac12 \sin(4x)$. Hence,

$\begin{align*} \int \sin(2x)\cos(2x) dx &= \int \cfrac12 \sin(4x)dx \\ &= -\cfrac18 \cos(4x)+C \\ &= \cfrac14 \sin^2(2x)+C \end{align*}$