Calculus/Integration/Solutions

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Integration of Polynomials[edit]

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[edit]

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[edit]

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}