Corrected exercises: integration tricks, Weierstrass, King's rule, Wallis (201-SN3-RE)

Let $u=a+b-x$, so $x=a+b-u$ and $dx=-du$. When $x=a$, $u=b$; when $x=b$, $u=a$. Then $\int_{a}^{b}f(x)\,dx=\int_{b}^{a}f(a+b-u)\left(-du\right)=\int_{a}^{b}f(a+b-u)\,du$. Since the name of the variable is a dummy, this is indeed the same integral. Geometrically, the graph has been flipped about the vertical line $x=\frac{a+b}{2}$, which does not change the area.

Here $a+b=\frac{\pi}{2}$, and $\sin\left(\frac{\pi}{2}-x\right)=\cos x$, $\cos\left(\frac{\pi}{2}-x\right)=\sin x$. The trick therefore turns $I$ exactly into $J$: $I=J$. Moreover $I+J=\int_{0}^{\pi/2}\dfrac{\sin x+\cos x}{\sin x+\cos x}\,dx=\int_{0}^{\pi/2}1\,dx=\dfrac{\pi}{2}$. From $I=J$ and $I+J=\frac{\pi}{2}$ we get $I=\dfrac{\pi}{4}$. No antiderivative was ever looked for: that is the whole point.

$\dfrac{1}{1+\tan^{\sqrt{2}}x}=\dfrac{\cos^{\sqrt{2}}x}{\cos^{\sqrt{2}}x+\sin^{\sqrt{2}}x}$. Calling this integral $I$, King's rule swaps $\sin$ and $\cos$ and gives $J=\int_{0}^{\pi/2}\dfrac{\sin^{\sqrt{2}}x}{\cos^{\sqrt{2}}x+\sin^{\sqrt{2}}x}dx$, with $I+J=\frac{\pi}{2}$ and $I=J$ by the same argument as in b. Hence $I=\dfrac{\pi}{4}$. The exponent disappeared because it never plays any role: all that matters is the SYMMETRY $\sin\leftrightarrow\cos$ of the interval. The answer would be the same with exponent 17 or $\pi$, which makes this integral unbeatable on a multiple-choice exam.

The trick with $x\mapsto\pi-x$ gives $K=\int_{0}^{\pi}\dfrac{\left(\pi-x\right)\sin x}{1+\cos^{2}x}\,dx$, since $\sin\left(\pi-x\right)=\sin x$ and $\cos^{2}\left(\pi-x\right)=\cos^{2}x$. Adding the two expressions: $2K=\pi\int_{0}^{\pi}\dfrac{\sin x}{1+\cos^{2}x}\,dx$. This last integral is done with $u=\cos x$, $du=-\sin x\,dx$: it equals $\int_{-1}^{1}\dfrac{du}{1+u^{2}}=\left[\arctan u\right]_{-1}^{1}=\dfrac{\pi}{4}+\dfrac{\pi}{4}=\dfrac{\pi}{2}$. Hence $2K=\pi\cdot\dfrac{\pi}{2}$ and $K=\dfrac{\pi^{2}}{4}\approx 2.467$. The factor $x$, which made the integral out of reach, has been eliminated by the trick.

Call this integral $I$. King's rule with $x\mapsto a-x$ gives $I=\int_{0}^{a}\dfrac{f(a-x)}{f(a-x)+f(x)}\,dx$. Adding the two expressions, the numerators recombine: $2I=\int_{0}^{a}\dfrac{f(x)+f(a-x)}{f(x)+f(a-x)}\,dx=\int_{0}^{a}1\,dx=a$, so $I=\dfrac{a}{2}$, WHATEVER $f$ is. Questions b and c are two special cases of it, with $a=\frac{\pi}{2}$. The shape to recognize: a quotient whose denominator is the sum of the numerator and its reflection.

With $t=\tan\frac{x}{2}$, the double-angle formulas give $\sin x=2\sin\frac{x}{2}\cos\frac{x}{2}$. Dividing top and bottom by $\cos^{2}\frac{x}{2}$: $\sin x=\dfrac{2\tan\frac{x}{2}}{\sec^{2}\frac{x}{2}}=\dfrac{2t}{1+t^{2}}$. Likewise $\cos x=\cos^{2}\frac{x}{2}-\sin^{2}\frac{x}{2}=\dfrac{1-\tan^{2}\frac{x}{2}}{1+\tan^{2}\frac{x}{2}}=\dfrac{1-t^{2}}{1+t^{2}}$. Finally $dt=\frac{1}{2}\sec^{2}\frac{x}{2}\,dx=\frac{1}{2}\left(1+t^{2}\right)dx$, hence $dx=\dfrac{2\,dt}{1+t^{2}}$.

$\displaystyle\int\dfrac{dx}{1+\sin x}=\int\dfrac{\frac{2\,dt}{1+t^{2}}}{1+\frac{2t}{1+t^{2}}}=\int\dfrac{2\,dt}{1+t^{2}+2t}=\int\dfrac{2\,dt}{\left(1+t\right)^{2}}=-\dfrac{2}{1+t}+C=-\dfrac{2}{1+\tan\frac{x}{2}}+C$. Check: differentiating, $\dfrac{d}{dx}\left[-\dfrac{2}{1+\tan\frac{x}{2}}\right]=\dfrac{2\cdot\frac{1}{2}\sec^{2}\frac{x}{2}}{\left(1+\tan\frac{x}{2}\right)^{2}}=\dfrac{1+t^{2}}{\left(1+t\right)^{2}}$. And $1+\sin x=1+\dfrac{2t}{1+t^{2}}=\dfrac{\left(1+t\right)^{2}}{1+t^{2}}$: the quotient obtained is indeed $\dfrac{1}{1+\sin x}$.

For $x\in\left[0,\frac{\pi}{2}\right]$, $t=\tan\frac{x}{2}$ goes from 0 to 1. $\displaystyle\int_{0}^{\pi/2}\dfrac{dx}{2+\cos x}=\int_{0}^{1}\dfrac{\frac{2\,dt}{1+t^{2}}}{2+\frac{1-t^{2}}{1+t^{2}}}=\int_{0}^{1}\dfrac{2\,dt}{2+2t^{2}+1-t^{2}}=\int_{0}^{1}\dfrac{2\,dt}{3+t^{2}}$. Hence $\dfrac{2}{\sqrt{3}}\left[\arctan\dfrac{t}{\sqrt{3}}\right]_{0}^{1}=\dfrac{2}{\sqrt{3}}\cdot\dfrac{\pi}{6}=\dfrac{\pi}{3\sqrt{3}}=\dfrac{\pi\sqrt{3}}{9}\approx 0.6046$.

The substitution requires $t=\tan\frac{x}{2}$ to be continuous, so $\frac{x}{2}$ must avoid $\frac{\pi}{2}$: it is only legitimate on an interval contained in $\left(-\pi,\pi\right)$, up to a shift by a multiple of $2\pi$. On $[0,2\pi]$, $t$ starts at 0, tends to $+\infty$ at $x=\pi$, reappears at $-\infty$ and climbs back to 0: blindly applying the antiderivative $\dfrac{2}{\sqrt{3}}\arctan\dfrac{t}{\sqrt{3}}$ gives $F\left(2\pi\right)-F\left(0\right)=0-0=0$. Yet the integrand $\dfrac{1}{2+\cos x}$ is everywhere greater than $\frac{1}{3}$: the integral is at least $\frac{2\pi}{3}$. The true value, obtained by splitting at $x=\pi$ and doubling by symmetry, is $\dfrac{2\pi}{\sqrt{3}}\approx 3.628$. Remember: a zero for the integral of a strictly positive function is not a computation error, it is a DOMAIN error.

These are Bioche's rules, which rest on the invariance of the expression $f(x)\,dx$. (1) If $f(-x)\left(-dx\right)=f(x)\,dx$, that is, if $f$ is odd, set $u=\cos x$. (2) If the expression is invariant under $x\mapsto\pi-x$, set $u=\sin x$. (3) If it is invariant under $x\mapsto\pi+x$, set $u=\tan x$. Weierstrass is only used when NONE of the three applies, as for $\dfrac{1}{2+\cos x}$: it is a last resort, not a first reflex, because it often produces painful partial fractions where a $u=\cos x$ would have been enough.

If $f$ is EVEN, $\int_{-a}^{a}f=2\int_{0}^{a}f$. If $f$ is ODD, $\int_{-a}^{a}f=0$. Proof of the odd case: $\int_{-a}^{a}f=\int_{-a}^{0}f+\int_{0}^{a}f$. In the first one, let $u=-x$: it becomes $\int_{0}^{a}f\left(-u\right)du=-\int_{0}^{a}f\left(u\right)du$. The sum is therefore zero. The hidden hypothesis, and it is the one behind question c, is that $f$ be INTEGRABLE on $[-a,a]$.

$\dfrac{x^{3}}{1+x^{4}}$ is odd (odd numerator, even denominator) and continuous: the integral is 0. For the second one, careful: $x^{5}$ is odd and $\sin x$ is odd, so their PRODUCT is EVEN. And $x^{2}$ is even. The integrand is therefore entirely even, and $\int_{-2}^{2}\left(x^{5}\sin x+x^{2}\right)dx=2\int_{0}^{2}\left(x^{5}\sin x+x^{2}\right)dx$, which has no reason to be zero; the term $x^{2}$ alone contributes $2\cdot\frac{8}{3}=\frac{16}{3}$. The classic error is to see $x^{5}$ and conclude “odd” without looking at the next factor.

The argument is wrong. The function $\frac{1}{x}$ is indeed odd, but it is NOT integrable on $[-1,1]$: it is not defined at 0, and the two halves $\int_{0}^{1}\frac{dx}{x}$ and $\int_{-1}^{0}\frac{dx}{x}$ diverge to $+\infty$ and $-\infty$ respectively. The integral does not exist; writing $\infty-\infty=0$ is not a computation. One can define a Cauchy principal value equal to 0, but it is a different object, and it must be named. Rule: parity never exempts you from checking continuity on the whole interval.

Let $u=\pi-x$ on $\left[\frac{\pi}{2},\pi\right]$: $du=-dx$, the bounds become $\frac{\pi}{2}$ and 0, and $\sin\left(\pi-u\right)=\sin u$. Hence $\int_{\pi/2}^{\pi}f\left(\sin x\right)dx=\int_{0}^{\pi/2}f\left(\sin u\right)du$: the two halves are equal, hence the factor 2. Indeed, the graph of $\sin$ is symmetric about the vertical line $x=\frac{\pi}{2}$. Application: $\int_{0}^{\pi}\sin^{4}x\,dx=2\int_{0}^{\pi/2}\sin^{4}x\,dx=2\cdot\dfrac{3\pi}{16}=\dfrac{3\pi}{8}$, the value $\frac{3\pi}{16}$ coming from the Wallis integrals of exercise 5.

Let $\varphi(a)=\int_{a}^{a+T}g$. Then $\varphi'(a)=g\left(a+T\right)-g\left(a\right)=0$ by periodicity, so $\varphi$ is constant and equals $\varphi(0)=\int_{0}^{T}g$. Application: $\left|\sin x\right|$ is $\pi$-periodic, and $\int_{0}^{\pi}\left|\sin x\right|dx=\int_{0}^{\pi}\sin x\,dx=\left[-\cos x\right]_{0}^{\pi}=2$. The interval $[0,10\pi]$ contains exactly 10 periods, so $\int_{0}^{10\pi}\left|\sin x\right|dx=10\cdot 2=20$. Without the absolute value, the same integral would be 0: it is the same pitfall as in exercise 9 of the set on polar coordinates.

$\cosh^{2}u-\sinh^{2}u=\dfrac{\left(e^{u}+e^{-u}\right)^{2}-\left(e^{u}-e^{-u}\right)^{2}}{4}=\dfrac{4e^{u}e^{-u}}{4}=1$. Differentiating the definitions: $\left(\sinh u\right)'=\dfrac{e^{u}+e^{-u}}{2}=\cosh u$ and $\left(\cosh u\right)'=\dfrac{e^{u}-e^{-u}}{2}=\sinh u$. Note the absence of a minus sign, unlike the trigonometric case: that is what makes these substitutions so comfortable.

Let $x=\sinh u=\dfrac{e^{u}-e^{-u}}{2}$ and $E=e^{u}$. Then $2x=E-\dfrac{1}{E}$, that is $E^{2}-2xE-1=0$. Solving gives $E=x\pm\sqrt{x^{2}+1}$, and since $E=e^{u}>0$ while $x-\sqrt{x^{2}+1}<0$, only the positive root works: $e^{u}=x+\sqrt{x^{2}+1}$, so $u=\ln\left(x+\sqrt{x^{2}+1}\right)$.

With $x=3\sinh u$, $dx=3\cosh u\,du$ and $\sqrt{x^{2}+9}=3\sqrt{\sinh^{2}u+1}=3\cosh u$, positive with no sign discussion. The integral becomes $\int\dfrac{3\cosh u}{3\cosh u}\,du=u+C=\sinh^{-1}\dfrac{x}{3}+C=\ln\left(x+\sqrt{x^{2}+9}\right)+C'$. Along the trigonometric route $x=3\tan\theta$, you get $\int\sec\theta\,d\theta=\ln\left|\sec\theta+\tan\theta\right|+C$, and then you must rebuild the right triangle to translate back $\sec\theta=\frac{\sqrt{x^{2}+9}}{3}$. Same result, two more steps, and an integral of secant to know by heart.

With $x=3\sinh u$: when $x=0$, $u=0$; when $x=3$, $\sinh u=1$, so $u_{1}=\ln\left(1+\sqrt{2}\right)$. The integral becomes $\int_{0}^{u_{1}}3\cosh u\cdot 3\cosh u\,du=9\int_{0}^{u_{1}}\cosh^{2}u\,du$. With $\cosh^{2}u=\dfrac{1+\cosh 2u}{2}$, an antiderivative is $\dfrac{u+\sinh u\cosh u}{2}$. At $u_{1}$, $\sinh u_{1}=1$ and $\cosh u_{1}=\sqrt{2}$, hence $\displaystyle\int_{0}^{3}\sqrt{x^{2}+9}\,dx=\dfrac{9}{2}\left[\ln\left(1+\sqrt{2}\right)+\sqrt{2}\right]\approx 10.330$.

For $\sqrt{x^{2}-a^{2}}$ we set $x=a\cosh u$, because $\cosh^{2}u-1=\sinh^{2}u$. But $\cosh u\geq 1$ for every real $u$: this substitution can only reach $x\geq a$. It is therefore reserved for the RIGHT branch of the domain $\left|x\right|\geq a$; for $x\leq -a$, set $x=-a\cosh u$, or handle the question by symmetry. Moreover $\sqrt{x^{2}-a^{2}}=a\left|\sinh u\right|$, and we choose $u\geq 0$ to remove the absolute value. It is the same precaution as for $x=a\sec\theta$, with a lighter computation behind it.

Let $u=\sin^{n-1}x$ and $dv=\sin x\,dx$, so $du=\left(n-1\right)\sin^{n-2}x\cos x\,dx$ and $v=-\cos x$. The boundary term $\left[-\sin^{n-1}x\cos x\right]_{0}^{\pi/2}$ vanishes at both bounds. What remains is $I_{n}=\left(n-1\right)\int_{0}^{\pi/2}\sin^{n-2}x\cos^{2}x\,dx$. Replacing $\cos^{2}x$ with $1-\sin^{2}x$: $I_{n}=\left(n-1\right)\left(I_{n-2}-I_{n}\right)$, that is $nI_{n}=\left(n-1\right)I_{n-2}$ and $I_{n}=\dfrac{n-1}{n}I_{n-2}$.

$I_{0}=\int_{0}^{\pi/2}dx=\dfrac{\pi}{2}$ and $I_{1}=\int_{0}^{\pi/2}\sin x\,dx=1$. Then $I_{2}=\frac{1}{2}I_{0}=\dfrac{\pi}{4}$, $I_{3}=\frac{2}{3}I_{1}=\dfrac{2}{3}$, $I_{4}=\frac{3}{4}I_{2}=\dfrac{3\pi}{16}$, $I_{5}=\frac{4}{5}I_{3}=\dfrac{8}{15}$, $I_{6}=\frac{5}{6}I_{4}=\dfrac{5\pi}{32}$.

For $n=2m$ even: $I_{2m}=\dfrac{2m-1}{2m}\cdot\dfrac{2m-3}{2m-2}\cdots\dfrac{1}{2}\cdot\dfrac{\pi}{2}$. For $n=2m+1$ odd: $I_{2m+1}=\dfrac{2m}{2m+1}\cdot\dfrac{2m-2}{2m-1}\cdots\dfrac{2}{3}\cdot 1$. The recurrence goes down two at a time: it ends at $I_{0}$ if $n$ is even, at $I_{1}$ if it is odd. Since $I_{0}=\frac{\pi}{2}$ contains $\pi$ and $I_{1}=1$ does not, only the even indices inherit $\pi$. The parity of the exponent therefore decides the nature of the result, which is a good way to check an answer: $\int_{0}^{\pi/2}\sin^{7}x\,dx$ cannot contain $\pi$.

The sequence $J_{n}=nI_{n}I_{n-1}$ satisfies $J_{n+1}=\left(n+1\right)I_{n+1}I_{n}=\left(n+1\right)\cdot\dfrac{n}{n+1}I_{n-1}\cdot I_{n}=nI_{n}I_{n-1}=J_{n}$: it is CONSTANT. Now $J_{1}=1\cdot I_{1}I_{0}=\dfrac{\pi}{2}$. Hence $nI_{n}I_{n-1}=\dfrac{\pi}{2}$, that is $I_{n}I_{n-1}=\dfrac{\pi}{2n}$. Check for $n=3$: $I_{3}I_{2}=\dfrac{2}{3}\cdot\dfrac{\pi}{4}=\dfrac{\pi}{6}$, and $\dfrac{\pi}{2\cdot 3}=\dfrac{\pi}{6}$. It is this identity, together with $\frac{I_{n}}{I_{n-1}}\to 1$, that gives the Wallis product $\dfrac{\pi}{2}=\prod_{m\geq 1}\dfrac{4m^{2}}{4m^{2}-1}$ and the asymptotic equivalent $I_{n}\sim\sqrt{\dfrac{\pi}{2n}}$.

Let $\Gamma_{n}=\int_{0}^{\infty}x^{n}e^{-x}dx$. By parts with $u=x^{n}$ and $dv=e^{-x}dx$: $\Gamma_{n}=\left[-x^{n}e^{-x}\right]_{0}^{\infty}+n\int_{0}^{\infty}x^{n-1}e^{-x}dx=n\Gamma_{n-1}$, the boundary term being zero because the exponential beats any power. Since $\Gamma_{0}=\int_{0}^{\infty}e^{-x}dx=1$, the recurrence gives $\Gamma_{n}=n!$. We have just proved that the factorial is an integral: it is the starting point of the Gamma function, which extends $n!$ to non-integer arguments.