# How to use u substitution for sin2x?

Jul 2, 2018

$\int \setminus \sin 2 x \setminus \mathrm{dx} = - \frac{1}{2} \cos 2 x + C$

#### Explanation:

We seek:

$\int \setminus \sin 2 x \setminus \mathrm{dx}$

Whilst we could perform a substitution, and in the early days of learning integration, this is perhaps the method used, the prefered method for such an integral is practice so that we can write the solution directly.

So, the preferred technique is to find an anti-derivative by differentiating a suitable function and then adjusting the function until we get a solution.

If we consider the likely candidate $\cos 2 x$ then using the chain rule we get:

$\frac{d}{\mathrm{dx}} \cos 2 x = - 2 \sin 2 x$

Hence, we have:

$- \setminus \int \setminus 2 \sin 2 x = \cos 2 x + c$

Thus we get:

$\int \setminus \sin 2 x = - \frac{1}{2} \cos 2 x + C$

Jul 2, 2018

$- \frac{1}{2} \cos \left(2 x\right) + C$

#### Explanation:

Given: $I = \int \sin 2 x \setminus \mathrm{dx}$.

Let $u = 2 x , \therefore \mathrm{du} = 2 \setminus \mathrm{dx} , \mathrm{dx} = \frac{\mathrm{du}}{2}$.

$\therefore I = \int \sin u \setminus \frac{\mathrm{du}}{2}$

$= \frac{1}{2} \int \sin u \setminus \mathrm{du}$

$= \frac{1}{2} \cdot - \cos u + C$

$= - \frac{1}{2} \cos u + C$

Inputting back $u = 2 x$, we get:

color(blue)(barul(|-1/2cos(2x)+C|)