How do you find the integral of # sin^2 (ax)#?

Question

Emma Aldridge · Accepted Answer

#int sin^2(ax) dx= ( ax-sin(ax)cos(ax) )/(2a)+C#

Use the trigonometric identity:

#sin^2(ax) = (1-cos(2ax))/2#

So:

#int sin^2(ax) dx= int (1-cos(2ax))/2dx#

#int sin^2(ax) dx= 1/2 int dx -1/2 int cos(2ax)dx#

#int sin^2(ax) dx= x/2 -1/(4a) int cos(2ax)d(2ax)#

#int sin^2(ax) dx= x/2 -1/(4a)sin(2ax) +C#

#int sin^2(ax) dx= x/2 -1/(2a)sin(ax)cos(ax) +C#

#int sin^2(ax) dx= ( ax-sin(ax)cos(ax) )/(2a)+C#

Emma Aldridge · Answer

undefined To find the integral of $\sin^2(ax)$, you can use trigonometric identities to rewrite it in terms of basic trigonometric functions, and then integrate. One of the most common identities for $\sin^2(x)$ is $\sin^2(x) = \frac{1}{2} - \frac{1}{2} \cos(2x)$. Applying this identity to $\sin^2(ax)$, we get:

$$
\sin^2(ax) = \frac{1}{2} - \frac{1}{2} \cos(2ax)
$$

Now, integrate both sides with respect to $x$:

$$
\int \sin^2(ax) \, dx = \int \left( \frac{1}{2} - \frac{1}{2} \cos(2ax) \right) \, dx
$$

$$
= \frac{1}{2} \int 1 \, dx - \frac{1}{2} \int \cos(2ax) \, dx
$$

The integral of $1$ with respect to $x$ is $x$, and the integral of $\cos(2ax)$ with respect to $x$ is $\frac{1}{2a} \sin(2ax)$ (using the chain rule). So, integrating both terms:

$$
= \frac{1}{2}x - \frac{1}{4a} \sin(2ax) + C
$$

where $C$ is the constant of integration. Therefore, the integral of $\sin^2(ax)$ is:

$$
\int \sin^2(ax) \, dx = \frac{1}{2}x - \frac{1}{4a} \sin(2ax) + C
$$