# Calculus/Integration/Solutions

## Integration of Polynomials

Evaluate the following:

1. $\int (x^{2}-2)^{2}\,dx$
{\begin{aligned}\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{aligned}}
{\begin{aligned}\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{aligned}}
2. $\int 8x^{3}\,dx$
{\begin{aligned}\int 8x^{3}dx&={\frac {8x^{4}}{4}}+C\\&=\mathbf {2x^{4}+C} \end{aligned}}
{\begin{aligned}\int 8x^{3}dx&={\frac {8x^{4}}{4}}+C\\&=\mathbf {2x^{4}+C} \end{aligned}}
3. $\int (4x^{2}+11x^{3})\,dx$
$\int (4x^{2}+11x^{3})dx=\mathbf {{\frac {4x^{3}}{3}}+{\frac {11x^{4}}{4}}+C}$
$\int (4x^{2}+11x^{3})dx=\mathbf {{\frac {4x^{3}}{3}}+{\frac {11x^{4}}{4}}+C}$
4. $\int (31x^{32}+4x^{3}-9x^{4})\,dx$
{\begin{aligned}\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{aligned}}
{\begin{aligned}\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{aligned}}
5. $\int 5x^{-2}\,dx$
{\begin{aligned}\int 5x^{-2}dx&={\frac {5x^{-1}}{-1}}+C\\&=\mathbf {-{\frac {5}{x}}+C} \end{aligned}}
{\begin{aligned}\int 5x^{-2}dx&={\frac {5x^{-1}}{-1}}+C\\&=\mathbf {-{\frac {5}{x}}+C} \end{aligned}}

## Indefinite Integration

Find the general antiderivative of the following:

6. $\int (\cos x+\sin x)\,dx$
$\int (\cos x+\sin x)\,dx=\mathbf {\sin x-\cos x+C}$
$\int (\cos x+\sin x)\,dx=\mathbf {\sin x-\cos x+C}$
7. $\int 3\sin x\,dx$
$\int 3\sin x\,dx=\mathbf {-3\cos(x)+C}$
$\int 3\sin x\,dx=\mathbf {-3\cos(x)+C}$
8. $\int (1+\tan ^{2}x)\,dx$
{\begin{aligned}\int (1+\tan ^{2}x)dx&=\int \sec ^{2}xdx\\&=\mathbf {\tan x+C} \end{aligned}}
{\begin{aligned}\int (1+\tan ^{2}x)dx&=\int \sec ^{2}xdx\\&=\mathbf {\tan x+C} \end{aligned}}
9. $\int (3x-\sec ^{2}x)\,dx$
$\int (3x-\sec ^{2}x)\,dx=\mathbf {{\frac {3x^{2}}{2}}-\tan x+C}$
$\int (3x-\sec ^{2}x)\,dx=\mathbf {{\frac {3x^{2}}{2}}-\tan x+C}$
10. $\int -e^{x}\,dx$
$\int -e^{x}\,dx=\mathbf {-e^{x}+C}$
$\int -e^{x}\,dx=\mathbf {-e^{x}+C}$
11. $\int 8e^{x}\,dx$
$\int 8e^{x}\,dx=\mathbf {8e^{x}+C}$
$\int 8e^{x}\,dx=\mathbf {8e^{x}+C}$
12. $\int {\frac {1}{7x}}\,dx$
$\int {\frac {1}{7x}}\,dx=\mathbf {{\frac {1}{7}}\ln |x|+C}$
$\int {\frac {1}{7x}}\,dx=\mathbf {{\frac {1}{7}}\ln |x|+C}$
13. $\int {\frac {1}{x^{2}+a^{2}}}\,dx$
Let
$x=a\tan \theta ;\qquad dx=a\sec ^{2}\theta d\theta$

Then

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

Then

{\begin{aligned}\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{aligned}}

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

(b)

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

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

${\frac {\sin ^{2}x}{2}}-(-{\frac {\cos ^{2}x}{2}})={\frac {\sin ^{2}x+\cos ^{2}x}{2}}={\frac {1}{2}}$
(a)
$u=\sin x;\qquad du=\cos xdx$
$v=\sin x;\qquad dv=\cos xdx$
${\begin{array}{ccc}\int \sin x\cos xdx=\sin ^{2}x-\int \sin x\cos xdx&\implies &2\int \sin x\cos xdx=\sin ^{2}x\\&\implies &\int \sin x\cos xdx=\mathbf {\frac {\sin ^{2}x}{2}} \end{array}}$

(b)

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

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

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