# Calculus/Integration/Solutions

## Integration of Polynomials

Evaluate the following:

1. $\int (x^2-2)^{2}\, dx$

\begin{align}\int(x^{2}-2)^{2}dx&=\int(x^{4}-4x^{2}+4)dx\\ &=\mathbf{\frac{x^{5}}{5}-\frac{4x^{3}}{3}+4x+C}\end{align}

2. $\int 8x^3\, dx$

\begin{align}\int8x^{3}dx&=\frac{8x^{4}}{4}+C\\ &=\mathbf{2x^{4}+C}\end{align}

3. $\int (4x^2+11x^3)\, dx$

$\int(4x^{2}+11x^{3})dx=\mathbf{\frac{4x^{3}}{3}+\frac{11x^{4}}{4}+C}$

4. $\int (31x^{32}+4x^3-9x^4) \,dx$

\begin{align}\int(31x^{32}+4x^{3}-9x^{4})dx&=\frac{31x^{33}}{33}+\frac{4x^{4}}{4}-\frac{9x^{5}}{5}+C\\ &=\mathbf{\frac{31x^{33}}{33}+x^{4}-\frac{9x^{5}}{5}+C}\end{align}

5. $\int 5x^{-2}\, dx$

\begin{align}\int5x^{-2}dx&=\frac{5x^{-1}}{-1}+C\\ &=\mathbf{-\frac{5}{x}+C}\end{align}

## Indefinite Integration

Find the general antiderivative of the following:

6. $\int (\cos x+\sin x)\, dx$

$\int (\cos x+\sin x)\, dx=\mathbf{\sin x-\cos x+C}$

7. $\int 3\sin x\, dx$

$\int 3\sin x\, dx=\mathbf{-3\cos(x)+C}$

8. $\int (1+\tan^2 x)\, dx$

\begin{align}\int(1+\tan^{2}x)dx&=\int\sec^{2}x dx\\ &=\mathbf{\tan x+C}\end{align}

9. $\int (3x-\sec^2 x)\, dx$

$\int (3x-\sec^2 x)\, dx=\mathbf{\frac{3x^{2}}{2}-\tan x+C}$

10. $\int -e^x\, dx$

$\int -e^x\, dx=\mathbf{-e^{x}+C}$

11. $\int 8e^x\, dx$

$\int 8e^x\, dx=\mathbf{8e^{x}+C}$

12. $\int \frac1{7x}\, dx$

$\int \frac1{7x}\, dx=\mathbf{\frac{1}{7}\ln|x|+C}$

13. $\int \frac1{x^2+a^2}\, dx$

Let

$x=a\tan\theta;\qquad dx=a\sec^{2}\theta d\theta$

Then

\begin{align}\int\frac{1}{x^{2}+a^{2}}dx&=\int\frac{a\sec^{2}\theta d\theta}{a^{2}(\tan^{2}\theta+1)}\\ &=\int\frac{\sec^{2}\theta d\theta}{a\sec^{2}\theta}\\ &=\frac{1}{a}\int d\theta\\ &=\frac{\theta}{a}+C\\ &=\mathbf{\frac{1}{a}\arctan\frac{x}{a}+C}\end{align}

## Integration by parts

14. Consider the integral $\int \sin(x) \cos(x)\,dx$. Find the integral in two different ways. (a) Integrate by parts with $u=\sin(x)$ and $v' =\cos(x)$. (b) Integrate by parts with $u=\cos(x)$ and $v' =\sin(x)$. Compare your answers. Are they the same?

(a)

$u=\sin x;\qquad du=\cos x dx$
$v=\sin x;\qquad dv=\cos x dx$
$\begin{array}{ccc} \int\sin x\cos x dx=\sin^{2}x-\int\sin x\cos x dx & \implies & 2\int\sin x\cos x dx=\sin^{2}x\\ & \implies & \int\sin x\cos x dx=\mathbf{\frac{\sin^{2}x}{2}} \end{array}$

(b)

$u=\cos x;\qquad du=-\sin x dx$
$v=-\cos x\qquad dv=\sin x dx$
$\begin{array}{ccc} \int\sin x\cos x dx=-\cos^{2}x-\int\sin x\cos x dx & \implies & 2\int\sin x\cos x dx=-\cos^{2}x\\ & \implies & \int\sin x\cos x dx=\mathbf{-\frac{\cos^{2}x}{2}} \end{array}$

Notice that the answers in parts (a) and (b) are not equal. However, since indefinite integrals include a constant term, we expect that the answers we found will differ by a constant. Indeed

$\frac{\sin^{2}x}{2}-(-\frac{\cos^{2}x}{2})=\frac{\sin^{2}x+\cos^{2}x}{2}=\frac{1}{2}$