Q. 68 - Calculus

Question

Solve each of the integrals in Exercises 63–68, where a, b, and c are real numbers with 

$$\begin{gathered} a\ne 0 , b\ne 0 , c>1, and c\ne 0. \\ \int \left(a\sin \left(bx\right)-c\right)dx. \end{gathered}$$

Step-by-Step Solution

Answer

The value of the integral is $-\frac{a}{b}\cos \left(bx\right) -cx +k.$

Step 1. Given Information.

Given is the integral:

$$\begin{gathered} \int \left(a\sin \left(bx\right)-c\right)dx, \\ a\ne 0 , b\ne 0 , c>1, and c\ne 0, \\ and a,b,and c are cons\tan ts. \end{gathered}$$

Step 2. Formula involved.

$$\begin{gathered} \int \sin \left(x\right)dx = -\cos x+k, \\ \int cdx = cx+k. \end{gathered}$$

Step 3. Solving the integral.

$$\begin{gathered} \int \left(a\sin \left(bx\right)-c\right)dx \\ = \int a\sin \left(bx\right)dx - \int cdx \\ = a\int \sin \left(bx\right)dx - cx+k \\ Let bx = t \\ bdx = dt \\ dx = \frac{dt}{b}, \\ Put the value of dx in the integral we get, \\ = a\int \sin t\frac{dt}{b} -cx+k \\ = \frac{a}{b}\int \sin t dt -cx+k \\ = -\frac{a}{b}\cos t -cx+k \\ = -\frac{a}{b}\cos \left(bx\right) -cx+k. \end{gathered}$$