Q. 57 - 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: $\tan \left(x\right) = \frac{\sin \left(x\right)}{\cos \left(x\right)}$).

$\int \left(\frac{1}{x}\cos x-\sin x\ln x\right)dx.$

Step-by-Step Solution

Answer

The value of the integral is $\ln \left(x\right)\cos \left(x\right)+c$.

Step 1. Given Information.

Given is an integral: $\int \left(\frac{1}{x}\sin x-\sin x\ln x\right)dx.$

Step 2. Formula involved.

$\int (udv+vdu) = \int d\left(uv\right) = uv$

Step 3. Solving the integral.

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