sinxcos2xdx

The integral of sin(x)cos(2x) with respect to x, denoted as $\int \sin(x)\cos(2x)dx$, is a common calculus problem involving trigonometric functions. To solve this integral, one can employ various substitution techniques or trigonometric identities.

One common approach is to use the double angle identity for cosine, $\cos(2x) = 1 - 2\sin^2(x)$. Substituting this into the integral gives $\int \sin(x)(1 - 2\sin^2(x))dx$. This can be expanded into $\int (\sin(x) - 2\sin^3(x))dx$. The integral of $\sin(x)$ is $-\cos(x)$. The integral of $\sin^3(x)$ can be found by rewriting it as $\sin(x)(1-\cos^2(x))$ and then using a substitution.

Alternatively, one could use the product-to-sum identity. The identity $\sin A \cos B = \frac{1}{2}[\sin(A+B) + \sin(A-B)]$ can be applied here with $A=x$ and $B=2x$. This yields $\sin(x)\cos(2x) = \frac{1}{2}[\sin(x+2x) + \sin(x-2x)] = \frac{1}{2}[\sin(3x) + \sin(-x)]$. Since $\sin(-x) = -\sin(x)$, this becomes $\frac{1}{2}[\sin(3x) - \sin(x)]$. Integrating this expression is straightforward: $\frac{1}{2} \int (\sin(3x) - \sin(x))dx$. The integral of $\sin(3x)$ is $-\frac{1}{3}\cos(3x)$, and the integral of $\sin(x)$ is $-\cos(x)$.

Both methods lead to the same result, which is a function involving cosine terms and an integration constant. The exact form of the solution may vary slightly depending on the method and the specific trigonometric identities used.