Test the convergence of integrate 0to ♾️ sin²xdx

Question

ScanSolve AI Tutor · Accepted Answer

To test the convergence of the integral $\int_{0}^{\infty} \sin^2(x)\,dx$ , we first rewrite the integrand using a trigonometric identity.

Rewrite $\sin^2(x)$ .
We use the identity $\sin^2(x) = \frac{1 - \cos(2x)}{2}$ .
Set up the improper integral as a limit.
$\int_{0}^{\infty} \sin^2(x)\,dx = \lim_{b \to \infty} \int_{0}^{b} \frac{1 - \cos(2x)}{2}\,dx$
Evaluate the definite integral.
$\int_{0}^{b} \frac{1 - \cos(2x)}{2}\,dx = \frac{1}{2} \int_{0}^{b} (1 - \cos(2x))\,dx$ $= \frac{1}{2} \left[ x - \frac{\sin(2x)}{2} \right]_{0}^{b}$ $= \frac{1}{2} \left[ \left( b - \frac{\sin(2b)}{2} \right) - \left( 0 - \frac{\sin(0)}{2} \right) \right]$ $= \frac{1}{2} \left[ b - \frac{\sin(2b)}{2} \right]$
Evaluate the limit.
$\lim_{b \to \infty} \frac{1}{2} \left[ b - \frac{\sin(2b)}{2} \right] = \lim_{b \to \infty} \left( \frac{b}{2} - \frac{\sin(2b)}{4} \right)$
As $b \to \infty$ , the term $\frac{b}{2}$ approaches $\infty$ . The term $\frac{\sin(2b)}{4}$ oscillates between $-\frac{1}{4}$ and $\frac{1}{4}$ .
Since the first term goes to infinity and the second term is bounded, the entire expression goes to infinity.
$\lim_{b \to \infty} \left( \frac{b}{2} - \frac{\sin(2b)}{4} \right) = \infty$
Since the limit does not exist (it goes to infinity), the integral diverges.

The integral $\int_{0}^{\infty} \sin^2(x)\,dx$ is divergent.
$\text{Divergent}$

Answer