Simplify the trigonometric expression $\sin^2x + \sin x \cos^2 x$.

1. Problem Description

Simplify the trigonometric expression $\sin^2x + \sin x \cos^2 x$.

2. Solution Steps

We are given the expression $\sin^2x + \sin x \cos^2 x$.

We can factor out $\sin x$ from both terms:

$$\sin x (\sin x + \cos^2 x)$$

We know the trigonometric identity:

$$\sin^2 x + \cos^2 x = 1$$

From this, we can express $\sin x$ as $1 - \cos^2 x$, but that's not helpful in this case. However, we can write $\sin x = 1 - \cos^2 x$, and substitute $\sin^2 x = 1-\cos^2 x$.

Then, the expression is $\sin x (\sin x + \cos^2 x)$. We know that $\sin x + \cos^2 x$ is not equal to one.

Instead we will use the identity $\sin^2 x + \cos^2 x = 1$.

Then we can write $\sin x (\sin x + \cos^2 x) = \sin x (\sin x + \cos^2 x)$.

We have

$\sin x (\sin x + \cos^2 x) = \sin x (\sin x) + \sin x (\cos^2 x) = \sin^2 x + \sin x \cos^2 x$

We have the expression:

$$\sin x(\sin x + \cos^2 x)$$

Since $\sin^2 x + \cos^2 x = 1$, we have $\sin x + \cos^2 x \neq 1$ unless $\sin x = \sin^2 x$, which happens only if $\sin x = 0$ or $\sin x = 1$.

We can rewrite the expression inside the parenthesis.

$$\sin x = 1 - \cos^2 x$$

Thus, $\sin x + \cos^2 x = 1 - \cos^2 x + \cos^2 x = 1$

$$\sin x(\sin x + \cos^2 x) = \sin x(1) = \sin x$$

Also we can proceed by using the identity $\sin^2x = \sin x \sin x = \sin x (1-cos^2 x) = \sin x - \sin x \cos^2 x$, but then we would have

$\sin x - \sin x \cos^2 x + \sin x \cos^2 x = \sin x$

3. Final Answer

$\sin x$