Integration of sin square x and sin 2x

Integration of sin square x can be calculated using trigonometric identities .Here is the formula for it

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

Integration of sin 2x can be calculated using integration by substitutions .Here is the formula for it

$$ \int \sin(2x) \, dx= \frac {-cos (2x)}{2} + C$

Proof of Integration of sin square x

To integrate $ \sin^2(x) $, we use a trigonometric identity to simplify the expression before integrating. The identity commonly used is the power-reduction formula or double angle identity

$$ \sin^2(x) = \frac{1 – \cos(2x)}{2} $$

Now, let’s integrate using this identity:

$$ \int \sin^2(x) \, dx = \int \frac{1 – \cos(2x)}{2} \, dx $$

This integral can be split into two simpler integrals:

$$ = \frac{1}{2} \int dx – \frac{1}{2} \int \cos(2x) \, dx $$

Now, integrate each part:

1. The integral of 1 with respect to $ x $ is $ x $.
2. The integral of $ \cos(2x) $ is $ \frac{\sin(2x)}{2} $.

Proof of this Integral $\int \cos(2x) dx= \frac{\sin(2x)}{2}$

Let t=2x
dt=2 dx
or
dx= dt/2
Therefore
$\int \cos(2x) \; dx== \frac {1}{2} \int cos t \; dt= \frac{\sin(2x)}{2} $

So, the integral becomes:

$$ = \frac{1}{2} x – \frac{1}{4} \sin(2x) + C $$

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

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

Proof of Integration of sin (2x)

We need to prove

$$ \int \sin(2x) \, dx= \frac {-cos (2x)}{2} + C$

Proof

Let t=2x
dt=2 dx
or
dx= dt/2
Therefore
$\int \sin(2x) \; dx= \frac {1}{2} \int sin(t)\; dt= – \frac{\cos(2x)}{2} $

Definite Integral of sin square x

To find the definite integral of $ \sin^2(x) $ over a specific interval, we use the same approach as with the indefinite integral, but we’ll apply the limits of integration at the end.

So, the definite integral of $ \sin^2(x) $ from ( a ) to ( b ) is:

\[ \int_a^b \sin^2(x) \, dx = \frac{1}{2} (b – a) + \frac{1}{4} (\sin(2a) – \sin(2b)) \]

Example

$$ \int_0^\pi \sin^2(x) \, dx $$

First, we use the power-reduction formula:

$$ \sin^2(x) = \frac{1 – \cos(2x)}{2} $$

Now, integrate over the interval $ [0, \pi] $:

$$ \int_0^\pi \frac{1 – \cos(2x)}{2} \, dx $$

$$ = \frac{1}{2} \int_0^\pi dx – \frac{1}{2} \int_0^\pi \cos(2x) \, dx $$

$$= \frac{1}{2} x \Big|_0^\pi – \frac{1}{4} \sin(2x) \Big|_0^\pi $$
$$=\frac{\pi}{2} + 0=\frac{\pi}{2}$ $

Therefore, the definite integral of $ \sin^2(x) $ from $ 0 $ to $ \pi $ is $ \frac{\pi}{2} $. This result represents the area under the curve of $ \sin^2(x) $ between $ x = 0 $ and $ x = \pi $.

Solved Examples

Question 1

$$
\int x \sin^2(x) \, dx
$$

Solution

To solve the integral

$$
\int x \sin^2(x) \, dx
$$

we’ll again use the integration by parts method. The formula for integration by parts is:

$$
\int u \, dv = uv – \int v \, du
$$

For this problem, let’s choose:

• $u = x$, which means $du = dx$,
• $dv = \sin^2(x) \, dx$, and we need to find $v$.

Now integral from above
$v= \int dv = \int \sin^2(x) \, dx = \frac{x}{2} – \frac{\sin(2x)}{4}$
Applying integration by parts with $u = x$ and $v = \frac{x}{2} – \frac{\sin(2x)}{4}$, we obtain:

$$
\int x \sin^2(x) \, dx = \frac{x^2}{4} – \frac{x \sin(2x)}{4} – \frac{\cos(2x)}{8} + C
$$