Question

Sin^2x dx integration

Answer 1

$\frac{x}{2} - \frac{\sin2x}{4} + C$

Answer 2

To integrate $\sin^2x \, dx$, we use the trigonometric identity that relates $\sin^2x$ to $\cos 2x$.

The identity is: $\cos 2x = 1 - 2\sin^2x$

Rearranging this identity to express $\sin^2x$: $2\sin^2x = 1 - \cos 2x$ $\sin^2x = \frac{1 - \cos 2x}{2}$

Now, substitute this into the integral: $\int \sin^2x \, dx = \int \frac{1 - \cos 2x}{2} \, dx$ $= \frac{1}{2} \int (1 - \cos 2x) \, dx$ We can split this into two separate integrals: $= \frac{1}{2} \left( \int 1 \, dx - \int \cos 2x \, dx \right)$ Integrate each term: The integral of $1$ with respect to $x$ is $x$. For the integral of $\cos 2x$, we can use a substitution. Let $u = 2x$, then $du = 2 \, dx$, which means $dx = \frac{1}{2} \, du$. $\int \cos 2x \, dx = \int \cos u \cdot \frac{1}{2} \, du = \frac{1}{2} \int \cos u \, du = \frac{1}{2} \sin u + C_1$ Substitute back $u = 2x$: $\int \cos 2x \, dx = \frac{1}{2} \sin 2x + C_1$ Now, combine these results: $\int \sin^2x \, dx = \frac{1}{2} \left( x - \frac{1}{2} \sin 2x \right) + C$ $= \frac{x}{2} - \frac{\sin 2x}{4} + C$ where $C$ is the constant of integration.

The core idea is to transform the $\sin^2x$ term into a form that is easier to integrate using a trigonometric identity. The identity $\sin^2x = \frac{1 - \cos 2x}{2}$ is crucial as it converts the squared trigonometric function into a linear term of a multiple angle, which can be integrated directly using standard integral formulas.