Q. 58 - Calculus

Question

Use integration formulas to solve each integral in Exercises 21–62. You may have to use algebra, educated guess- and- check, and/or recognize an integrand as the result of a product, quotient, or chain rule calculation. Check each of your answers by differentiating. (Hint for Exercise 54: $\tan \left(x\right) = \frac{\sin \left(x\right)}{\cos \left(x\right)}$).

$\int \frac{2x\sin x-(x^2+1)\cos x}{{\sin }^{2}x}dx.$

Step-by-Step Solution

Answer

The value of the integral is $\frac{\left(x^2+1\right)}{\sin x}+c$.

Step 1. Given Information.

Given is a integral: $\int \frac{2x\sin x-(x^2+1)\cos x}{{\sin }^{2}x}dx.$

Step 2. Formula involved.

$\int \frac{udv-vdu}{u^2} = \frac{v}{u}.$

Step 3. Solving the integral.

$$\begin{gathered} \int \frac{2x\sin x-(x^2+1)\cos x}{{\sin }^{2}x}dx \\ = \int \frac{\sin xd(x^2+1)-(x^2+1)d\left(\sin x\right)}{{\sin }^{2}x}dx \\ = \frac{(x^2+1)}{\sin x}+c. \end{gathered}$$