]> Exercise 3

# Definitions and Basics, Exercise 3

## Question

Calculate the following integrals:

1. $\int {\mathrm{tan}}^{2}x\text{ }\text{d}x$

2. $\int \frac{\text{d}x}{{\mathrm{tan}}^{2}x}\text{ }$

3. $\int \frac{\text{d}x}{1-{x}^{4}}\text{ }$
Hint for the last case: $\left(1-{x}^{4}\right)=\left(1+{x}^{2}\right)\left(1-{x}^{2}\right)$

## Reminder

 $\left\{\begin{array}{l}\int \mathrm{cos}x\text{ }\text{d}x=\mathrm{sin}x+C\\ \int \mathrm{sin}x\text{ }\text{d}x=-\mathrm{cos}x+C\\ \int \left(1+{\mathrm{tan}}^{2}x\right)\text{d}x=\int \frac{\text{d}x}{{\mathrm{cos}}^{2}x}=\mathrm{tan}x+C\\ \int \left(1+{\mathrm{cot}}^{2}x\right)\text{d}x=\int \frac{\text{d}x}{{\mathrm{sin}}^{2}x}=-\mathrm{cot}x+C\end{array}\right\}$ (2.1.1.14)
 $\left\{\begin{array}{l}\int \frac{\text{d}x}{\sqrt{1-{x}^{2}}}=\left(\text{or}\right)\left\{\begin{array}{l}±\text{asin}\text{ }x+C\\ \mp \text{acos}\text{ }x+C\end{array}\\ \int \frac{\text{d}x}{1+{x}^{2}}=\left(\text{or}\right)\left\{\begin{array}{l}+\text{atan}\text{ }x+C\\ -\text{acot}\text{ }x+C\end{array}\end{array}\right\}$ (2.1.1.15)
 $\left\{\begin{array}{l}\int \frac{\text{d}x}{\sqrt{1+{x}^{2}}}=\text{asinh}\text{ }x+C\\ \int \frac{\text{d}x}{\sqrt{{x}^{2}-1}}=\mp \text{acosh}\text{ }x+C\\ \int \frac{\text{d}x}{1-{x}^{2}}=\text{atanh}\text{ }x+C\text{ }\left(|x|<1\right)\\ \int \frac{\text{d}x}{1-{x}^{2}}=\text{acoth}\text{ }x+C\text{ }\left(|x|>1\right)\end{array}\right\}$ (2.1.1.18)

## Parts 1-2

Solution of question 1

1. Among the integrals known so far the $\int {\mathrm{tan}}^{2}x\text{ }\text{d}x$ is included in (2.1.1.14): $\int \left(1+{\mathrm{tan}}^{2}x\right)\text{d}x=\mathrm{tan}x+C$
2. as a consequence: $\int {\mathrm{tan}}^{2}x\text{ }\text{d}x=\mathrm{tan}x-\int \text{d}x=\mathrm{tan}x-x+C$

## Parts 3-5

Solution of question 2

1. Among the integrals known so far the $\int \frac{\text{d}x}{{\mathrm{tan}}^{2}x}\text{ }$ is not included, but $\frac{1}{\mathrm{tan}x}=\mathrm{cot}x$

2. and therefore (2.1.1.14) could be used: $\int \text{d}x\left(1+{\mathrm{cot}}^{2}x\right)=\int \text{d}x\left(1+\frac{1}{{\mathrm{tan}}^{2}x}\right)=-\frac{1}{\mathrm{tan}x}+C$

3. As a consequence: $\int \frac{\text{d}x}{{\mathrm{tan}}^{2}x}\text{ }=-\frac{1}{\mathrm{tan}x}-\int \text{d}x=-\left(\frac{1}{\mathrm{tan}x}+x\right)+C$

## Parts 6-9

Solution of question 3

1. The integral $\int \frac{\text{d}x}{1-{x}^{4}}\text{ }$ does not appear among the known integrals,
but according to the hint one have to look for the expressions (1+x2) and (1−x2) and luckily they are in (2.1.1.15) and (2.1.1.18).
2. We have to seek for a possible solution of $\frac{1}{1-{x}^{4}}=\frac{A}{1+{x}^{2}}+\frac{B}{1-{x}^{2}}$
3. Indeed we obtain $\frac{A}{1+{x}^{2}}+\frac{B}{1-{x}^{2}}=\frac{\left(A+B\right)+\left(-A+B\right){x}^{2}}{1-{x}^{4}}$ and the requirement is fulfilled with $A=B=\frac{1}{2}$
4. Finally we obtain $\begin{array}{l}\int \frac{\text{d}x}{1-{x}^{4}}=\frac{1}{2}\int \frac{\text{d}x}{1+{x}^{2}}+\frac{1}{2}\int \frac{\text{d}x}{1-{x}^{2}}=\\ =\frac{\text{atan}x}{\text{2}}+\left\{\begin{array}{l}\frac{1}{2}\text{atanh}\text{ }x\text{ }\text{for}\left(|x|<1\right)\\ \frac{1}{2}\text{acoth}\text{ }x\text{ }\text{for}\left(|x|>1\right)\end{array}\right\}+C\end{array}$

## Score

By questions:

Question 1 gives credit of 2 points.
Question 2 gives credit of 3 points.
Question 3 gives credit of 5 points.

By parts:

Parts 1,2,3,4,5,6,8,9 are worth 1 point each.
Part 7 is worth 2 points.