Chapter 1 Proofs: Integration Formulas

Proof

Let $F(x) = \dfrac{x^{n+1}}{n+1}$. Since $n \neq -1$, we have $n+1 \neq 0$, so this expression is well-defined. Differentiating gives

$$F'(x) = \frac{1}{n+1} \cdot (n+1)\,x^{(n+1)-1} = x^n$$

Therefore $F(x)$ is an antiderivative of $x^n$, and

$$\int x^n \, dx = \frac{x^{n+1}}{n+1} + C$$

$\square$

Proof

We consider two cases for $x \neq 0$.

Case 1: $x > 0$.

Since $|x| = x$, we have $\ln|x| = \ln x$. By the differentiation formula for the natural logarithm,

$$\frac{d}{dx}\ln x = \frac{1}{x}$$

Case 2: $x < 0$.

Since $|x| = -x$, we have $\ln|x| = \ln(-x)$. By the chain rule,

$$\frac{d}{dx}\ln(-x) = \frac{1}{-x} \cdot (-1) = \frac{1}{x}$$

In both cases, $(\ln|x|)' = 1/x$ holds. Therefore,

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

$\square$

Proof

By the differentiation formula for the exponential function,

$$\frac{d}{dx}e^x = e^x$$

That is, $e^x$ is its own antiderivative. Therefore,

$$\int e^x \, dx = e^x + C$$

$\square$

Proof

Rewrite $a^x = e^{x \ln a}$. Let $F(x) = \dfrac{a^x}{\ln a} = \dfrac{e^{x \ln a}}{\ln a}$. By the chain rule,

$$F'(x) = \frac{1}{\ln a} \cdot e^{x \ln a} \cdot \ln a = e^{x \ln a} = a^x$$

The condition $a \neq 1$ is needed to ensure $\ln a \neq 0$ (so that the denominator is nonzero). Therefore,

$$\int a^x \, dx = \frac{a^x}{\ln a} + C$$

$\square$

Proof

Differentiate the right-hand side.

$$\frac{d}{dx}(-\cos x + C) = -(-\sin x) = \sin x$$

This equals the integrand. $\square$

Proof

Differentiate the right-hand side.

$$\frac{d}{dx}(\sin x + C) = \cos x$$

This equals the integrand. $\square$

Proof

Apply the quotient rule to $\tan x = \dfrac{\sin x}{\cos x}$.

$$(\tan x)' = \frac{(\sin x)'\cos x - \sin x\,(\cos x)'}{\cos^2 x} = \frac{\cos x \cdot \cos x - \sin x \cdot (-\sin x)}{\cos^2 x}$$ $$= \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x$$

Therefore $\tan x$ is an antiderivative of $\sec^2 x$, and

$$\int \sec^2 x \, dx = \tan x + C$$

$\square$

Proof

Apply the quotient rule to $\cot x = \dfrac{\cos x}{\sin x}$.

$$(\cot x)' = \frac{(\cos x)'\sin x - \cos x\,(\sin x)'}{\sin^2 x} = \frac{-\sin x \cdot \sin x - \cos x \cdot \cos x}{\sin^2 x}$$ $$= \frac{-(\sin^2 x + \cos^2 x)}{\sin^2 x} = \frac{-1}{\sin^2 x} = -\csc^2 x$$

Therefore $(-\cot x)' = \csc^2 x$, and

$$\int \csc^2 x \, dx = -\cot x + C$$

$\square$

Proof

We find $(\arcsin x)'$ using the inverse function differentiation rule. Let $y = \arcsin x$, so that $x = \sin y$ ($-\pi/2 \le y \le \pi/2$). Then

$$\frac{dx}{dy} = \cos y$$

By the inverse function differentiation rule,

$$\frac{dy}{dx} = \frac{1}{\cos y}$$

Since $\cos y \ge 0$ in the range $-\pi/2 \le y \le \pi/2$, and $\sin^2 y + \cos^2 y = 1$, we have

$$\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}$$

Therefore,

$$(\arcsin x)' = \frac{1}{\sqrt{1 - x^2}}$$

This equals the integrand, so the formula holds. $\square$

Proof

Let $y = \arccos x$, so that $x = \cos y$ ($0 \le y \le \pi$). By the inverse function differentiation rule,

$$\frac{dy}{dx} = \frac{1}{-\sin y}$$

Since $\sin y \ge 0$ in the range $0 \le y \le \pi$, and $\sin^2 y + \cos^2 y = 1$, we have

$$\sin y = \sqrt{1 - \cos^2 y} = \sqrt{1 - x^2}$$

Therefore,

$$(\arccos x)' = \frac{-1}{\sqrt{1 - x^2}}$$

This equals the integrand. $\square$

Proof

Let $y = \arctan x$, so that $x = \tan y$ ($-\pi/2 < y < \pi/2$). Then

$$\frac{dx}{dy} = \sec^2 y = 1 + \tan^2 y = 1 + x^2$$

By the inverse function differentiation rule,

$$(\arctan x)' = \frac{dy}{dx} = \frac{1}{1 + x^2}$$

This equals the integrand, so the formula holds. $\square$

Proof

Let $y = \operatorname{arcsec} x$ ($x > 1$), so that $x = \sec y$ ($0 \le y < \pi/2$). The derivative of $\sec y$ is

$$(\sec y)' = \sec y \tan y$$

By the inverse function differentiation rule,

$$\frac{dy}{dx} = \frac{1}{\sec y \tan y}$$

Here $\sec y = x$, and since $\tan^2 y = \sec^2 y - 1 = x^2 - 1$, we have $\tan y = \sqrt{x^2 - 1}$ (since $\tan y \ge 0$ for $0 \le y < \pi/2$). Therefore,

$$(\operatorname{arcsec} x)' = \frac{1}{x\sqrt{x^2-1}} \quad (x > 1)$$

For $x < -1$, taking the principal value of $\operatorname{arcsec}$ in $[0, \pi/2) \cup (\pi/2, \pi]$, we have $\sec y = x$ and $\tan y < 0$, so

$$(\operatorname{arcsec} x)' = \frac{1}{\sec y \tan y} = \frac{1}{x \cdot (-\sqrt{x^2-1})} = \frac{-1}{x\sqrt{x^2-1}} = \frac{1}{|x|\sqrt{x^2-1}}$$

(since $x < 0$ implies $-1/x = 1/|x|$). In both cases,

$$(\operatorname{arcsec} x)' = \frac{1}{|x|\sqrt{x^2-1}} \quad (|x| > 1)$$

This equals the integrand. $\square$

Proof

Differentiating from the definition $\cosh x = (e^x + e^{-x})/2$, we get

$$(\cosh x)' = \frac{d}{dx}\frac{e^x + e^{-x}}{2} = \frac{e^x - e^{-x}}{2} = \sinh x$$

Therefore $\cosh x$ is an antiderivative of $\sinh x$. $\square$

Proof

Differentiating from the definition $\sinh x = (e^x - e^{-x})/2$, we get

$$(\sinh x)' = \frac{d}{dx}\frac{e^x - e^{-x}}{2} = \frac{e^x + e^{-x}}{2} = \cosh x$$

Therefore $\sinh x$ is an antiderivative of $\cosh x$. $\square$

Proof

Apply the quotient rule to $\tanh x = \dfrac{\sinh x}{\cosh x}$.

$$(\tanh x)' = \frac{(\sinh x)'\cosh x - \sinh x\,(\cosh x)'}{\cosh^2 x} = \frac{\cosh^2 x - \sinh^2 x}{\cosh^2 x}$$

By the hyperbolic identity $\cosh^2 x - \sinh^2 x = 1$,

$$(\tanh x)' = \frac{1}{\cosh^2 x} = \operatorname{sech}^2 x$$

Therefore $\tanh x$ is an antiderivative of $\operatorname{sech}^2 x$, and the formula holds. $\square$

Proof

Apply the quotient rule to $\coth x = \dfrac{\cosh x}{\sinh x}$.

$$(\coth x)' = \frac{(\cosh x)'\sinh x - \cosh x\,(\sinh x)'}{\sinh^2 x} = \frac{\sinh^2 x - \cosh^2 x}{\sinh^2 x}$$ $$= \frac{-(\cosh^2 x - \sinh^2 x)}{\sinh^2 x} = \frac{-1}{\sinh^2 x} = -\operatorname{csch}^2 x$$

Therefore $(-\coth x)' = \operatorname{csch}^2 x$, and

$$\int \operatorname{csch}^2 x \, dx = -\coth x + C$$

$\square$

Proof

Method 1 (Verification): Differentiate $F(x) = \ln(x + \sqrt{x^2+1})$. Let $u(x) = x + \sqrt{x^2+1}$. Then

$$u'(x) = 1 + \frac{x}{\sqrt{x^2+1}} = \frac{\sqrt{x^2+1} + x}{\sqrt{x^2+1}}$$

Therefore,

$$F'(x) = \frac{u'(x)}{u(x)} = \frac{\dfrac{\sqrt{x^2+1} + x}{\sqrt{x^2+1}}}{x + \sqrt{x^2+1}} = \frac{1}{\sqrt{x^2+1}}$$

This equals the integrand.

Method 2 (Derivation): Substitute $x = \sinh t$. Then $dx = \cosh t\,dt$ and $\sqrt{x^2+1} = \sqrt{\sinh^2 t + 1} = \cosh t$, so

$$\int \frac{1}{\sqrt{x^2+1}}\,dx = \int \frac{\cosh t}{\cosh t}\,dt = \int dt = t + C = \operatorname{arcsinh} x + C$$

$\square$

Proof

Method 1 (Verification): Differentiate $F(x) = \ln(x + \sqrt{x^2-1})$. Let $u(x) = x + \sqrt{x^2-1}$. Then

$$u'(x) = 1 + \frac{x}{\sqrt{x^2-1}} = \frac{\sqrt{x^2-1} + x}{\sqrt{x^2-1}}$$

Therefore,

$$F'(x) = \frac{u'(x)}{u(x)} = \frac{\dfrac{\sqrt{x^2-1} + x}{\sqrt{x^2-1}}}{x + \sqrt{x^2-1}} = \frac{1}{\sqrt{x^2-1}}$$

This equals the integrand.

Method 2 (Derivation): For $x > 1$, substitute $x = \cosh t$ ($t > 0$). Then $dx = \sinh t\,dt$ and $\sqrt{x^2-1} = \sqrt{\cosh^2 t - 1} = \sinh t$ (since $\sinh t > 0$ for $t > 0$), so

$$\int \frac{1}{\sqrt{x^2-1}}\,dx = \int \frac{\sinh t}{\sinh t}\,dt = \int dt = t + C = \operatorname{arccosh} x + C$$

$\square$

Proof

Method 1 (Verification): Differentiate $F(x) = \dfrac{1}{2}\ln\!\left|\dfrac{1+x}{1-x}\right| = \dfrac{1}{2}\bigl(\ln|1+x| - \ln|1-x|\bigr)$.

$$F'(x) = \frac{1}{2}\left(\frac{1}{1+x} - \frac{-1}{1-x}\right) = \frac{1}{2}\left(\frac{1}{1+x} + \frac{1}{1-x}\right)$$

Finding a common denominator,

$$F'(x) = \frac{1}{2} \cdot \frac{(1-x) + (1+x)}{(1+x)(1-x)} = \frac{1}{2} \cdot \frac{2}{1-x^2} = \frac{1}{1-x^2}$$

This equals the integrand.

Method 2 (Derivation): Use partial fraction decomposition.

$$\frac{1}{1-x^2} = \frac{1}{(1+x)(1-x)} = \frac{1}{2}\left(\frac{1}{1+x} + \frac{1}{1-x}\right)$$

Integrating each term,

$$\int \frac{1}{1-x^2}\,dx = \frac{1}{2}\bigl(\ln|1+x| - \ln|1-x|\bigr) + C = \frac{1}{2}\ln\!\left|\frac{1+x}{1-x}\right| + C$$

For $|x| < 1$, this equals $\operatorname{arctanh} x + C$. $\square$