Integrating trigonometric, exponential and reciprocal functions questions with answers

Questions

1. Question 1sin, cos and sec²

   Find $\int 4\sec^{2} (2x - 1) \, dx$, where $x$ is in radians.

   Show the worked solution

   1. Step 1: Recall the integral of sec² (check by differentiating it back)$\text{The integral of sec² is tan.}$
   2. Step 2: Integrate, then divide by the coefficient of x (to undo the chain rule)$\int 4\sec^{2} (2x - 1) \, dx = \dfrac{4}{2}\tan (2x - 1) + c$$= 2\tan (2x - 1) + c$
2. Question 2sin, cos and sec²

   Find $\int 4\sin 6x \, dx$, where $x$ is in radians.

   Show the worked solution

   1. Step 1: Recall the integral of sin (check by differentiating it back)$\text{The integral of sin is −cos.}$
   2. Step 2: Integrate, then divide by the coefficient of x (to undo the chain rule)$\int 4\sin 6x \, dx = -\dfrac{4}{6}\cos 6x + c$$= -\dfrac{2}{3}\cos 6x + c$
3. Question 3Exponentials

   Find $\int 6e^{-4x + 1} \, dx$.

   Show the worked solution

   1. Step 1: Recall the integral of e to the power x (check by differentiating it back)$\text{The integral of e to the power x is e to the power x itself.}$
   2. Step 2: Integrate, then divide by the coefficient of x (to undo the chain rule)$\int 6e^{-4x + 1} \, dx = \dfrac{6}{(-4)}e^{-4x + 1} + c$$= -\dfrac{3}{2}e^{-4x + 1} + c$
4. Question 4Exponentials

   Find $\int 3e^{-x} \, dx$.

   Show the worked solution

   1. Step 1: Recall the integral of e to the power x (check by differentiating it back)$\text{The integral of e to the power x is e to the power x itself.}$
   2. Step 2: Integrate, then divide by the coefficient of x (to undo the chain rule)$\int 3e^{-x} \, dx = \dfrac{3}{(-1)}e^{-x} + c$$= -3e^{-x} + c$
5. Question 51 over ax + b

   Find $\int \dfrac{3}{x + 4} \, dx$, where $x > -4$.

   Show the worked solution

   1. Step 1: Recall the integral of 1/x (check by differentiating it back)$\text{The integral of 1/x is ln x.}$
   2. Step 2: Integrate, then divide by the coefficient of x (to undo the chain rule)$\int \dfrac{3}{x + 4} \, dx = \dfrac{3}{1}\ln(x + 4) + c$$= 3\ln(x + 4) + c$
6. Question 61 over ax + b

   Find $\int \dfrac{1}{7x + 3} \, dx$, where $x > -\dfrac{3}{7}$.

   Show the worked solution

   1. Step 1: Recall the integral of 1/x (check by differentiating it back)$\text{The integral of 1/x is ln x.}$
   2. Step 2: Integrate, then divide by the coefficient of x (to undo the chain rule)$\int \dfrac{1}{7x + 3} \, dx = \dfrac{1}{7}\ln(7x + 3) + c$$= \dfrac{1}{7}\ln(7x + 3) + c$