6.5: Miscellaneous Integration Methods

The integration methods presented so far are considered “standard,” meaning every calculus student should know them. This section will discuss a few additional methods, some more common than others. One such method is the Leibniz integral rule for “differentiation under the integral sign.”⁵ This powerful and useful method is best explained with a simple example.

$$\frac{d}{\dalpha} \int e^{\alpha x}\;\dx ~=~ \int \frac{d}{\dalpha}\,(e^{\alpha x})~\dx ~=~ \int x\,e^{\alpha x}\;\dx \nonumber$$

However, differentiating the right side of the formula ([eqn:diffinteax]) shows that

$$\frac{d}{\dalpha} \int e^{\alpha x}\;\dx ~=~ \frac{d}{\dalpha} \left(\tfrac{1}{\alpha}\,e^{\alpha x} ~+~ C\right) ~=~ \frac{\alpha\,\left(x\,e^{\alpha x}\right) ~-~ 1\,\cdot\,e^{\alpha x}}{\alpha^2} ~=~ \tfrac{1}{\alpha}\,x\,e^{\alpha x} ~-~ \tfrac{1}{\alpha^2}\,e^{\alpha x} \nonumber$$

Thus,

$$\int x\,e^{\alpha x}\;\dx ~=~ \tfrac{1}{\alpha}\,x\,e^{\alpha x} ~-~ \tfrac{1}{\alpha^2}\,e^{\alpha x} ~+~ C \nonumber$$

which can be verified via integration by parts with the tabular method:

$$\int x\,e^{\alpha x}\;\dx ~=~ \tfrac{1}{\alpha}\,x\,e^{\alpha x} ~-~ \tfrac{1}{\alpha^2}\,e^{\alpha x} ~+~ C\quad\checkmark \nonumber$$

What was actually done in the above example? A known integral,

$$\int e^{\alpha x}\;\dx ~=~ \tfrac{1}{\alpha}\,e^{\alpha x} ~+~ C ~, \nonumber$$

was differentiated with respect to $\alpha$ via the Leibniz rule to produce a new integral,

$$\int x\,e^{\alpha x}\;\dx ~=~ \tfrac{1}{\alpha}\,x\,e^{\alpha x} ~-~ \tfrac{1}{\alpha^2}\,e^{\alpha x} ~+~ C ~, \nonumber$$

with the constant $\alpha$ treated temporarily—only during the differentiation—as a variable. In general, that is how the Leibniz rule is used. Typically this means if you want to evaluate a certain integral with the Leibniz rule, then you “work backwards” to figure out which integral you need to differentiate with respect to some constant (e.g. $\alpha$) in the integrand.

Example $\PageIndex{1}$: intleibniz1

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Solution

Use the Leibniz rule to evaluate $~\displaystyle\int \frac{\dx}{(1 + x^2)^2}~$.

Solution: By formula ([eqn:atanint]) in Section 5.4,

$$\int\,\frac{\dx}{a^2 + x^2} ~=~ \tfrac{1}{a}\,\tan^{-1}\left(\tfrac{x}{a}\right) ~+~ C \nonumber$$

for any constant $a > 0$. So differentiate both sides with respect to $a$:

\[\begin{aligned} \frac{d}{\da}\,\int\,\frac{\dx}{a^2 + x^2} ~&=~ \frac{d}{\da}\,\left(\tfrac{1}{a}\,\tan^{-1}\left(\tfrac{x}{a}\right) ~+~ C\right)\

\[6pt] \int\,\frac{d}{\da}\,\left(\frac{1}{a^2 + x^2}\right)~\dx ~&=~ -\tfrac{1}{a^2}\,\tan^{-1}\left(\tfrac{x}{a}\right) ~+~ \tfrac{1}{a}\,\cdot\,\frac{1}{1 + \left(\tfrac{x}{a}\right)^2}\,\cdot\,-\tfrac{x}{a^2}\

\[6pt] \int -\frac{2a}{(a^2 + x^2)^2}\,\dx ~&=~ -\tfrac{1}{a^2}\,\tan^{-1}\left(\tfrac{x}{a}\right) ~-~ \frac{x}{a\,(a^2 + x^2)}\

$$6pt] \int \frac{\dx}{(a^2 + x^2)^2} ~&=~ \tfrac{1}{2a^3}\,\tan^{-1}\left(\tfrac{x}{a}\right) ~+~ \frac{x}{2a^2\,(a^2 + x^2)} ~+~ C\end{aligned} \nonumber$$

That general formula is useful in itself. In particular, for $a=1$,

$$\int \frac{\dx}{(1 + x^2)^2} ~=~ \tfrac{1}{2}\,\tan^{-1} x ~+~ \frac{x}{2\,(1 + x^2)} ~+~ C ~, \nonumber$$

which agrees with the result from Example

Example $\PageIndex{1}$: trigsub2

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Solution

in Section 6.3.
Notice that there was no generic constant (e.g. $a$ or $\alpha$) in the statement of the problem. When that happens, you will need to figure out where the constant should be in order to use the Leibniz rule.

You can also use differentiation under the integral sign to evaluate definite integrals.

Example $\PageIndex{1}$: intexpx2

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Solution

Show that $~\displaystyle\int_0^{\infty} e^{-x^2} \,\dx ~=~ \tfrac{1}{2}\sqrt{\pi}~$.

Solution: Let $I = \int_0^{\infty} e^{-x^2} \,\dx$. The integral is convergent, since by Exercise [exer:exple1px] in Section 4.4, for all $x$

$$e^{x^2} ~\ge~ 1 ~+~ x^2 \quad\Rightarrow\quad 0 ~\le~ e^{-x^2} ~\le~ \frac{1}{1 + x^2} \nonumber$$

implies $I$ is convergent by the Comparison Test, since $\int_0^{\infty} \frac{1}{1 + x^2}\,\dx$ is convergent (and equals $\tfrac{1}{2}\pi$) by Example

Example $\PageIndex{1}$: improper5

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Solution

in Section 5.5. For $\alpha \ge 0$, define

$$\phi(\alpha) ~=~ \int_0^{\infty} \,\frac{\alpha\,e^{-\alpha^2 x^2}}{1 + x^2} \,\dx ~. \nonumber$$

Then clearly $\phi(0) = 0$, and differentiating under the integral sign shows

$$\phi'(\alpha) ~=~ \int_0^{\infty} \,\frac{-2\alpha^2 e^{-\alpha^2 x^2} + e^{-\alpha^2 x^2}}{1 + x^2}~\dx \qquad\Rightarrow\qquad \phi'(0) ~=~ \int_0^{\infty} \frac{\dx}{1 + x^2} ~=~ \tfrac{1}{2}\pi ~. \nonumber$$

The substitution $y = \alpha x$, so that $\dy = \alpha\,\dx$, shows $\phi(\alpha)$ can be written as

$$\phi(\alpha) ~=~ \int_0^{\infty} \,\frac{e^{-y^2}}{1 + \left(\tfrac{y}{\alpha}\right)^2} \,\dy \qquad\Rightarrow\qquad 0 ~\le~ \lim_{\alpha \to \infty}~ \phi(\alpha) ~\le~ I ~<~ \infty ~. \nonumber$$

Also, for $\alpha > 0$,

\[\begin{aligned} \frac{d}{\dalpha}\,\left(\frac{1}{\alpha}\,e^{-\alpha^2}\,\phi(\alpha)\right) ~&=~ \frac{d}{\dalpha}\,\int_0^{\infty} \,\frac{e^{-\alpha^2 (1+x^2)}}{1 + x^2} \,\dx ~=~ \int_0^{\infty} \,\frac{-2\alpha\,(1+x^2)\, e^{-\alpha^2 (1+x^2)}}{1 + x^2}~\dx\

\[6pt] &=~ -2\alpha\, e^{-\alpha^2}\,\int_0^{\infty} e^{-\alpha^2 x^2}~\dx \quad\text{, now substitute $u = \alpha x$ and $\du = \alpha \dx$ to get}\

\[6pt] &=~ -2\alpha\, e^{-\alpha^2}\,\frac{1}{\alpha}\,\int_0^{\infty} e^{-u^2} \,\du ~=~ -2\, e^{-\alpha^2}\,I \quad\text{, and so integrating both sides yields}\

$$6pt] \int_0^{\infty} \frac{d}{\dalpha}\,\left(\frac{1}{\alpha}\,e^{-\alpha^2}\,\phi(\alpha)\right)~\dalpha ~&=~ -2I\,\int_0^{\infty} e^{-\alpha^2} \,\dalpha ~=~ -2I^2 ~.\end{aligned} \nonumber$$

However, by the Fundamental Theorem of Calculus.

\[\begin{aligned} \int_0^{\infty} \frac{d}{\dalpha}\,\left(\frac{1}{\alpha}\,e^{-\alpha^2}\,\phi(\alpha)\right)~\dalpha ~&=~ \frac{1}{\alpha}\,e^{-\alpha^2}\,\phi(\alpha)~\Biggr|_0^{\infty} ~=~ \left(\lim_{\alpha \to \infty}~\frac{\phi(\alpha)}{\alpha \,e^{\alpha^2}}\right) ~-~ \left(\lim_{\alpha \to 0}~\frac{\phi(\alpha)}{\alpha\,e^{\alpha^2}}\right)\

\[6pt] &=~ 0 ~-~ \left(\lim_{\alpha \to 0}~\frac{\phi(\alpha)}{\alpha\,e^{\alpha^2}}\right) ~\to~ \frac{0}{0} \quad\text{, so by L'H\^{o}pital's Rule}\

$$6pt] &=~ -\lim_{\alpha \to 0}~\frac{\phi'(\alpha)}{e^{\alpha^2} + 2\alpha^2\,e^{\alpha^2}} ~=~ -\frac{\phi'(0)}{1+0} ~=~ -\tfrac{1}{2}\pi ~.\end{aligned} \nonumber$$

Thus,

$$-2I^2 ~=~ -\tfrac{1}{2}\pi \qquad\Rightarrow\qquad I ~=~ \tfrac{1}{2}\sqrt{\pi} \nonumber$$

which is the desired result.

One immediate consequence of Example

Example $\PageIndex{1}$: intexpx2

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Solution

is that

$$\int_{-\infty}^{\infty} e^{-x^2} \,\dx ~=~ \sqrt{\pi} \nonumber$$

since $e^{-x^2}$ is an even function. The following example shows another consequence, as well as how useful substitutions can be in writing integrals in a different form.

Example $\PageIndex{1}$: intgamma1

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Solution

Show that the Gamma function $\Gamma\,(t)$ can be written as

$$\Gamma\,(t) ~=~ 2\,\int_0^{\infty} y^{2t-1} \, e^{-y^2} ~\dy \quad\text{for all $t > 0$,} \nonumber$$

and that $\Gamma\,\left(\tfrac{1}{2}\right) ~=~ \sqrt{\pi}$.

Solution: Let $x = y^2$, so that $\dx = 2y\;\dy$. Then $x=0~\Rightarrow~y=0~$ and $x=\infty~\Rightarrow~y=\infty$, so

\[\Gamma\,(t) ~=~ \int_0^{\infty} x^{t-1} \, e^{-x} ~\dx ~=~ \int_0^{\infty} (y^2)^{t-1}\,e^{-y^2}~2y~\dy\

$$6pt] ~=~ 2\,\int_0^{\infty} y^{2t-1} \, e^{-y^2} ~\dy ~. \nonumber$$

In this form, with the help of Example

Example $\PageIndex{1}$: intexpx2

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Solution

it is now easy to evaluate $\Gamma\,\left(\tfrac{1}{2}\right)$:

$$\Gamma\,\left(\tfrac{1}{2}\right) ~=~ 2\,\int_0^{\infty} y^{1-1} \, e^{-y^2} ~\dy ~=~ 2\,\int_0^{\infty} e^{-y^2}~\dy ~=~ 2\,\left(\tfrac{1}{2}\sqrt{\pi}\right) ~=~ \sqrt{\pi} \nonumber$$

$$\label{eqn:betagamma} B(x,y) ~=~ \frac{\Gamma\,(x)\;\Gamma\,(y)}{\Gamma\,(x+y)} \qquad\text{for all $x > 0$ and $y > 0$.}$$

Example $\PageIndex{1}$: intbeta1

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Solution

Show that the Beta function $B(x,y)$ can be written as

$$B(x,y) ~=~ \int_0^{\infty} \frac{u^{x-1}}{(1+u)^{x+y}}~\du ~. \nonumber$$

Solution: Let $u=\frac{t}{1-t}$, so that $t=\frac{u}{1+u}$, $1-t=\frac{1}{1+u}$, and $\dt = \frac{\du}{(1+u)^2}$. Then $t=0~\Rightarrow~u=0$ and $t=1~\Rightarrow~u=\infty$, so

$$B(x,y) ~=~ \int_0^1 t^{x-1}\,(1-t)^{y-1}\,\dt ~=~ \int_0^{\infty} \left(\frac{u}{1+u}\right)^{x-1}\;\left(\frac{1}{1+u}\right)^{y-1} \frac{\du}{(1+u)^2} ~=~ \int_0^{\infty} \frac{u^{x-1}}{(1+u)^{x+y}}~\du ~. \nonumber$$

Another application of substitutions in integrals is in the evaluation of fractional derivatives. Recall from Section 1.6 that the zero-th derivative of a function is just the function itself, and that derivatives of order $n$ are well-defined for integer values $n \ge 1$. It turns out that derivatives of fractional orders—e.g. 1/2—can be defined, with the Riemann-Louiville definition being the most common:

Example $\PageIndex{1}$: halfderivx

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Solution

Calculate $~\dfrac{d^{1/2}}{\dx^{1/2}}\,(x)~$.

Solution: Here $\alpha = \frac{1}{2}$ and $f(x)=x$, so that

$$\frac{d^{1/2}}{\dx^{1/2}}\,(x) ~=~ \frac{1}{\Gamma\,(1-1/2)}\;\ddx\,\int_0^x \frac{t}{(x-t)^{1/2}}\,\dt ~=~ \frac{1}{\sqrt{\pi}}\;\ddx\,\int_0^x \frac{t~\dt}{\sqrt{x-t}} \nonumber$$

since $\Gamma\,\left(\tfrac{1}{2}\right) ~=~ \sqrt{\pi}$ by Example

Example $\PageIndex{1}$: intgamma1

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Solution

. Use the substitution $u=\sqrt{x-t}$, so that $t=x-u^2$ and $\dt=-2u\,\du$. Then $t=0~\Rightarrow~u=\sqrt{x}~$ and $t=x~\Rightarrow~u=0$, so

\[\begin{aligned} \frac{d^{1/2}}{\dx^{1/2}}\,(x) ~&=~ \frac{1}{\sqrt{\pi}}\;\ddx\,\int_{\sqrt{x}}^0 \frac{(x-u^2)\,(-2u~\du)}{u} ~=~ \frac{2}{\sqrt{\pi}}\;\ddx\,\int_0^{\sqrt{x}} (x ~-~ u^2)~\du\

\[6pt] &=~ \frac{2}{\sqrt{\pi}}\;\ddx\,\left(xu ~-~ \tfrac{1}{3}\,u^3\right)~\Biggr|_{u=0}^{u=\sqrt{x}} ~=~ \frac{2}{\sqrt{\pi}}\;\ddx\,\left(x^{3/2} ~-~ \tfrac{1}{3}\,x^{3/2}\right)\

$$6pt] &=~ \frac{2}{\sqrt{\pi}}\;\ddx\,\left(\tfrac{2}{3}\,x^{3/2}\right) ~=~ \frac{2}{\sqrt{\pi}}\,\sqrt{x}\end{aligned} \nonumber$$

$$\frac{d^{n+\alpha}}{\dx^{n+\alpha}}\,f(x) ~=~ \frac{d^{\alpha}}{\dx^{\alpha}}\,\left(\frac{d^{n}}{\dx^{n}}\,f(x)\right) \nonumber$$

Recall from Section 6.3 that the trigonometric substitution $x=r\,\cos\,\theta$—or its sister substitution $x=r\,\sin\,\theta$—was motivated by trying to find the area of a circle of radius $r$. To simplify matters, let $r=1$ so that points on the unit circle can be identified with the angle $\theta$ via that substitution, with $\theta$ as shown in Figure [fig:circle2](a) below.

Figure [fig:circle2](b) shows a different identification of points on the unit circle—by slope. This will be the basis for a half-angle substitution for evaluating certain integrals.

Let $A$ be the point $(-1,0)$, then for any other point $P$ on the unit circle draw a line from $A$ through $P$ until it intersects the line $x=1$, as shown in Figure [fig:circle3] below:

From geometry you know that the inscribed angle that the line $\overline{AP}$ makes with the $x$-axis is half the measure of the central angle $\theta$. So the slope of $\overline{AP}$ is the tangent of that angle: $\tan\,\frac{1}{2}\theta = \frac{t}{1} = t$, which is measured along the $y$-axis and can take any real value. Each point on the unit circle—except $A$—can be identified with that slope $t$. Figure [fig:circle3] shows only positive slopes—reflect the picture about the $x$-axis for negative slopes. The figure shows that

$$\sin\,\tfrac{1}{2}\theta ~=~ \frac{t}{\sqrt{1+t^2}} \qquad\text{and}\qquad \cos\,\tfrac{1}{2}\theta ~=~ \frac{1}{\sqrt{1+t^2}} \nonumber$$

so that by the double-angle identities for sine and cosine,

$$\sin\,\theta ~=~ 2\,\sin\,\tfrac{1}{2}\theta\,\cos\,\tfrac{1}{2}\theta ~=~ 2\,\frac{t}{\sqrt{1+t^2}}\,\frac{1}{\sqrt{1+t^2}} ~=~ \frac{2t}{1+t^2} \nonumber$$

and

$$\cos\,\theta ~=~ \cos^2 \tfrac{1}{2}\theta ~-~ \sin^2 \tfrac{1}{2}\theta ~=~ \frac{1}{1+t^2} ~-~ \frac{t^2}{1+t^2} ~=~ \frac{1-t^2}{1+t^2} ~. \nonumber$$

Since $\theta = 2\,\tan^{-1} \,t$, then

$$\dtheta ~=~ d\,\left(2\,\tan^{-1} t\right) ~=~ \frac{2\,\dt}{1+t^2} ~. \nonumber$$

Below is a summary of the substitution: The half-angle substitution thus turns rational functions of $\sin\,\theta$ and $\cos\,\theta$ into rational functions of $t$, which can be integrated using partial fractions or another method.

Example $\PageIndex{1}$: inthalfangle1

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Solution

Evaluate $~\displaystyle\int \frac{\dtheta}{1 \;+\; \sin\,\theta \;+\; \cos\,\theta}$.

Solution: Using $t = \tan\,\tfrac{1}{2}\theta$, the denominator of the integrand is

$$1 ~+~ \sin\,\theta ~+~ \cos\,\theta ~=~ \frac{1+t^2}{1+t^2} ~+~ \frac{2t}{1+t^2} ~+~ \frac{1-t^2}{1+t^2} ~=~ \frac{2t + 2}{1+t^2} \nonumber$$

so that

\[\begin{aligned} \int \frac{\dtheta}{1 \;+\; \sin\,\theta \;+\; \cos\,\theta} ~&=~ \mathop{\mathlarger{\mathlarger{\int}}} \frac{\frac{2\,\dt}{1+t^2}}{\frac{2t + 2}{1+t^2}} ~=~ \int \frac{\dt}{t+1}\

$$6pt] &=~ \ln\,\abs{t+1} ~+~ C\\ &=~ \ln\,\Abs{\tan\,\tfrac{1}{2}\theta \;+\;1} ~+~ C\end{aligned} \nonumber$$

Example $\PageIndex{1}$: inthalfangle2

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Solution

Evaluate $~\displaystyle\int \frac{\dtheta}{3\,\sin\,\theta \;+\; 4\,\cos\,\theta}~$.

Solution: Using $t = \tan\,\tfrac{1}{2}\theta$, the integral becomes

\[\begin{aligned} \int \frac{\dtheta}{3\,\sin\,\theta \;+\;4\,\cos\,\theta} ~&=~ \mathop{\mathlarger{\mathlarger{\int}}} \frac{\frac{2\,\dt}{1+t^2}}{3\,\frac{2t}{1+t^2} \;+\; 4\,\frac{1-t^2}{1+t^2}} ~=~ \int \frac{-1}{2t^2 - 3t - 2}\,\dt\

$$6pt] &=~ \int \frac{-1}{(2t+1)\,(t-2)}\,\dt ~=~ \int \left(\frac{A}{2t+1} ~+~ \frac{B}{t-2}\right)\,\dt\end{aligned} \nonumber$$

where

$$\begin{aligned} {3} \text{coefficient of $t$}&: \quad & A ~+~ 2B ~&=~ 0 \quad\Rightarrow\quad A ~=~ -2B\\ \text{constant term}&: & -2A ~+~ B ~&=~ -1 \quad\Rightarrow\quad 4B ~+~ B ~=~ -1 \quad\Rightarrow\quad B ~=~ -\frac{1}{5} ~~\text{and}~~ A ~=~ \frac{2}{5}\end{aligned} \nonumber$$

Thus,

\[\begin{aligned} \int \frac{\dtheta}{3\,\sin\,\theta \;+\;4\,\cos\,\theta} ~&=~ \int \left(\frac{\frac{2}{5}}{2t+1} ~+~ \frac{-\frac{1}{5}}{t-2}\right)\,\dt ~=~ \frac{1}{5}\,\ln\,\abs{2t+1} ~-~ \frac{1}{5}\,\ln\,\abs{t-2} ~+~ C\

$$4pt] &=~ \frac{1}{5}\,\ln\,\Abs{2\,\tan\,\tfrac{1}{2}\theta \;+\; 1} ~-~ \frac{1}{5}\,\ln\,\Abs{\tan\,\tfrac{1}{2}\theta \;-\; 2} ~+~ C\end{aligned} \nonumber$$

By the half-angle substitution $t = \tan\,\tfrac{1}{2}\theta$,

$$\frac{\sin\,\theta}{1 \;+\; \cos\,\theta} ~=~ \frac{\dfrac{2t}{1+t^2}}{\dfrac{1+t^2}{1+t^2} + \dfrac{1-t^2}{1+t^2}} ~=~ \frac{\dfrac{2t}{1+t^2}}{\dfrac{2}{1+t^2}} ~=~ t \nonumber$$

which yields the useful half-angle identities:⁹

Example $\PageIndex{1}$: inthalfangle3

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Solution

Evaluate $~\displaystyle\int \frac{\sin\,\theta}{1 \;+\; \cos\,\theta}\,\dtheta~$.

Solution: Though you could use the half-angle substitution $t = \tan\,\tfrac{1}{2}\theta$, it is easier to use the half-angle identity ([eqn:halftan1]) directly, since

$$\int \frac{\sin\,\theta}{1 \;+\; \cos\,\theta}\,\dtheta ~=~ \int \tan\,\tfrac{1}{2}\theta~\dtheta ~=~ 2\,\ln\,\Abs{\sec\,\tfrac{1}{2}\theta} ~+~ C \nonumber$$

by formula ([eqn:inttanu]) in Section 6.3.

[sec6dot5]

For Exercises 1-12, evaluate the given integral.

4

$\displaystyle\int \frac{1 \;-\; 2\,\cos\,\theta}{\sin\,\theta}\;\dtheta$

$\displaystyle\int \frac{\dtheta}{3 \;-\; 5\,\sin\,\theta}$

$\displaystyle\int \frac{\dtheta}{2 \;-\; \sin\,\theta}$

$\displaystyle\int \frac{\dtheta}{4 \;+\; \sin\,\theta}$

4

$\displaystyle\int \frac{\sin\,\theta}{2 \;-\; \sin\,\theta}\;\dtheta$

$\displaystyle\int \frac{\dtheta}{5 \;-\; 3\,\cos\,\theta}$

$\displaystyle\int \frac{\dtheta}{1 \;+\; \sin\,\theta \;-\; \cos\,\theta}$

$\displaystyle\int \frac{\dtheta}{1 \;-\; \sin\,\theta \;+\; \cos\,\theta}$

4

$\displaystyle\int \frac{\cot\,\theta}{1 \;+\; \sin\,\theta}\;\dtheta$

$\displaystyle\int \frac{1 \;-\; \cos\,\theta}{3\,\sin\,\theta}\;\dtheta$

$\displaystyle\int_{-\infty}^{\infty} e^{-x^2/2}\,\dx$

$\displaystyle\int_{-\infty}^{\infty} x^2 \,e^{-x^6}\,\dx$

Consider the integral $~\displaystyle\int \frac{\sin\,\theta}{1 \;+\; \cos\,\theta}\,\dtheta~$ from Example

Example $\PageIndex{1}$: inthalfangle3

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Solution

.

1. Evaluate the integral using the substitution $u=1 + \cos\,\theta$.
2. Evaluate the integral using the half-angle substitution $t = \tan\,\tfrac{1}{2}\theta$.
3. Show that the answers from parts (a) and (b) are equivalent to the result from Example

   Example $\PageIndex{1}$: inthalfangle3

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   Solution

   .

[[1.]]

Evaluate the integral $~\displaystyle\int \frac{\dtheta}{3\,\sin\,\theta \;+\; 4\,\cos\,\theta}~$ from Example

Example $\PageIndex{1}$: inthalfangle2

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Solution

by noting that

\[\begin{aligned} \int \frac{\dtheta}{3\,\sin\,\theta \;+\; 4\,\cos\,\theta} ~&=~ \int \frac{\dtheta}{5\,\left(\frac{3}{5}\,\sin\,\theta \;+\; \frac{4}{5}\,\cos\,\theta\right)}\

\[5pt] &=~ \int \frac{\dtheta}{5\,\left(\cos\,\phi\;\sin\,\theta \;+\; \sin\,\phi\;\cos\,\theta\right)}\

$$5pt] &=~ \int \frac{\dtheta}{5\,\sin\,(\theta + \phi)} ~=~ \frac{1}{5}\,\int \csc\,(\theta + \phi)~\dtheta\end{aligned} \nonumber$$

by the sine addition formula, where $\phi$ is the angle in the right triangle shown above. Complete the integration and show that your answer is equivalent to the result from Example

Example $\PageIndex{1}$: inthalfangle2

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Solution

. [[1.]]

Show directly from the definition of the Beta function that $B(x,y) = B(y,x)$ for all $x > 0$ and $y > 0$.

[exer:betatrig] Show that the Beta function $B(x,y)$ can be written as

$$B(x,y) ~=~ \int_0^{\pi/2} 2\,\sin^{2x-1}(\theta)~\cos^{2y-1}(\theta)~\dtheta \qquad\text{for all $x > 0$ and $y > 0$.} \nonumber$$

[exer:intsinmcosn] Use Exercise [exer:betatrig] and formula ([eqn:betagamma]) to show that

$$\int_0^{\pi/2} \sin^{m}\theta~\cos^{n}\theta~\dtheta ~=~ \frac{\Gamma\,\left(\dfrac{m+1}{2}\right) \; \Gamma\,\left(\dfrac{n+1}{2}\right)}{2\,\Gamma\,\left(\dfrac{m+n}{2} + 1\right)} \qquad\text{for all $m > -1$ and $n > -1$.} \nonumber$$

Use Exercise [exer:gamma] from Section 6.1, as well as Exercise [exer:intsinmcosn] above, to show that for $m=1$, $2$, $3$, $\ldots$,

$$\int_0^{\pi/2} \sin^{2m}\theta~\dtheta ~=~ \frac{\sqrt{\pi}\;\Gamma\,\left(m + \frac{1}{2}\right)}{2\,(m!)} \qquad\text{and}\qquad \int_0^{\pi/2} \sin^{2m+1}\theta~\dtheta ~=~ \frac{\sqrt{\pi}\;(m!)}{2\,\Gamma\,\left(m + \frac{3}{2}\right)} ~. \nonumber$$

2

Show that $~\displaystyle\int_0^{\infty} \dfrac{\ln\,x}{1 + x^2}\,\dx ~=~ 0$.

Show that $~\displaystyle\int_0^{\infty} \dfrac{x^a}{a^x}\,\dx ~=~ \dfrac{\Gamma\,(a+1)}{(\ln\,a)^{a+1}}~$ for $a > 1$.

Use the result from Example

Example $\PageIndex{1}$: halfderivx

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Solution

to show that

$$\frac{d^{1/2}}{\dx^{1/2}}\,\left(\frac{d^{1/2}}{\dx^{1/2}}\,(x)\right) ~=~ 1 ~=~ \ddx\,(x) ~. \nonumber$$

2

Calculate $~\dfrac{d^{1/2}}{\dx^{1/2}}\,(c)~$ for all constants $c$.

Calculate $~\dfrac{d^{1/3}}{\dx^{1/3}}\,(x)~$.

Show that $~\displaystyle\int_0^1 \dfrac{1}{\sqrt{1 - x^n}}\,\dx ~=~ \tfrac{1}{n}\,B\left(\tfrac{1}{n},\tfrac{1}{2}\right)~$ for $n \ge 1$.

Show that the Gamma function $\Gamma\,(t)$ can be written as

$$\Gamma\,(t) ~=~ p^t\,\int_0^{\infty} u^{t-1} \,e^{-pu}~\du \quad\text{for all $t > 0$ and $p > 0$.} \nonumber$$

Show that the Gamma function $\Gamma\,(t)$ can be written as

$$\Gamma\,(t) ~=~ \int_0^1 \left(\ln\,\left(\frac{1}{u}\right)\right)^{t-1}\,\du \quad\text{for all $t > 0$.} \nonumber$$

Using the result from Exercise [exer:eaxtrigbx] in Section 6.1 that

$$\int e^{ax}\,\cos\,bx~\dx ~=~ \frac{e^{ax}\,(a\,\cos\,bx ~+~ b\,\sin\,bx)}{a^2 + b^2} \nonumber$$

for all constants $a$ and $b \ne 0$, differentiate under the integral sign to show that for all $\alpha > 0$

$$\int_0^{\infty} x\,e^{-x} \sin\,\alpha x~\dx ~=~ \frac{2 \alpha}{(1 + \alpha^2)^2} ~. \nonumber$$

[[1.]]

Use the Leibniz rule and formula ([eqn:sqrta2u2tan]) from Section 6.3 to show that for all $a > 0$,

$$\int \frac{\dx}{\sqrt{a^2 + x^2}} ~=~ \ln\;\Abs{x + \sqrt{a^2 + x^2}\,} ~+~ C ~. \nonumber$$

Use Example

Example $\PageIndex{1}$: intbeta1

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Solution

to show that the Beta function satisfies the relation

$$B(x,1-x) ~=~ \int_0^1 \,\frac{t^{-x} \;+\; t^{x-1}}{1 + t}\,\dt \quad\text{for all $0 < x < 1$.} \nonumber$$

(Hint: First use a substitution to show that $\displaystyle\int_0^{\infty} \dfrac{u^{x-1}}{1 + u}\,\du = \displaystyle\int_0^{\infty} \dfrac{t^{-x}}{1 + t}\,\dt$.)

Show that for all $a > -1$,

$$\int_0^{\pi/2} \frac{\dtheta}{1 \;+\; a\,\sin^2 \theta} ~=~ \frac{\pi}{2\,\sqrt{1+a}} ~. \nonumber$$