Question

$\int {\tan }^2(3x+5)dx$ is equal to (where $C$ is the constant of integration)

• A. $\frac{1}{3x+5}\tan (3x+5)-x+C$
• B. $\frac{1}{3}\tan (3x+5)+x+C$
• C. $\frac{1}{3}\tan (3x+5)-x+C$
• D. $\frac{1}{3x+5}\tan (3x+5)+x+C$

Answer 1

The correct option is C $\frac{1}{3}\tan (3x+5)-x+C$

$I=\int {\tan }^2(3x+5)dx=\int \left({\sec }^2\right(3x+5)-1)dx=\int {\sec }^2(3x+5)dx-\int dx\{\because \int {\sec }^2(ax+b)dx=\frac{1}{a}\tan (ax+b)+C,a\ne 0\}=\frac{1}{3}\tan (3x+5)-x+C$