Question

evaluate the integral using any appropriate algebraic method or trigonometric identity. (int\frac{10}{1 - cos(5x)}dx=square)

Explanation:

Step1: Use double - angle formula

We know that $1-\cos(2\theta)=2\sin^{2}\theta$. Let $\theta = \frac{5x}{2}$, then $1 - \cos(5x)=2\sin^{2}(\frac{5x}{2})$. The integral becomes $\int\frac{10}{2\sin^{2}(\frac{5x}{2})}dx = 5\int\csc^{2}(\frac{5x}{2})dx$.

Step2: Use integral formula

The integral of $\csc^{2}(u)$ is $-\cot(u)+C$. Let $u=\frac{5x}{2}$, then $du=\frac{5}{2}dx$. So $5\int\csc^{2}(\frac{5x}{2})dx=5\times\frac{-2}{5}\cot(\frac{5x}{2})+C=- 2\cot(\frac{5x}{2})+C$.

Answer:

$-2\cot(\frac{5x}{2})+C$