Question

$\int \frac{sin^2x}{(1+cosx)}dx = ?$

Answer 1

$x - \sin x + C$

Answer 2

To evaluate the integral $\int \frac{\sin^2x}{(1+\cos x)}dx$, we use trigonometric identities.

The fundamental trigonometric identity is $\sin^2x + \cos^2x = 1$, which implies $\sin^2x = 1 - \cos^2x$. Substitute this into the integral: $\int \frac{\sin^2x}{(1+\cos x)}dx = \int \frac{1 - \cos^2x}{(1+\cos x)}dx$ Next, we use the algebraic identity $a^2 - b^2 = (a-b)(a+b)$. Here, $a=1$ and $b=\cos x$. So, $1 - \cos^2x = (1 - \cos x)(1 + \cos x)$. Substitute this expression back into the integral: $\int \frac{(1 - \cos x)(1 + \cos x)}{(1+\cos x)}dx$ Assuming $1+\cos x \neq 0$ (i.e., $x \neq (2n+1)\pi$ for integer $n$), we can cancel the term $(1+\cos x)$ from the numerator and the denominator: $\int (1 - \cos x)dx$ Now, we integrate term by term: $\int 1 dx - \int \cos x dx$ The integral of $1$ with respect to $x$ is $x$. The integral of $\cos x$ with respect to $x$ is $\sin x$. Therefore, the result of the integration is: $x - \sin x + C$ where C is the constant of integration.

Explanation of the solution: The integral is simplified by using the identity $\sin^2x = 1 - \cos^2x$. This expression is then factored as a difference of squares, $(1-\cos x)(1+\cos x)$. The $(1+\cos x)$ term cancels out, leaving a simpler integral $\int (1-\cos x)dx$. Integrating term by term yields $x - \sin x + C$.