Integrals of tan x, cot x, sec x and cosec x

Corollary III of section 3.2.1 ($\int\frac{f'(x)}{f(x)}dx=\log|f(x)|+c$) immediately settles the integrals of $\tan x$, $\cot x$, $\sec x$ and $\csc x$.

$\int\tan x\,dx$. $\int\frac{\sin x}{\cos x}\,dx=-\int\frac{-\sin x}{\cos x}\,dx=-\log|\cos x|+c=\log|\sec x|+c$ (since $\frac{d}{dx}\cos x=-\sin x$).

$\int\cot(5x-4)\,dx$. $\int\frac{5\cos(5x-4)}{5\sin(5x-4)}\,dx=\frac15\log|\sin(5x-4)|+c$ (Corollary I applied to the basic $\cot$ result).

$\int\sec x\,dx$. Multiply and divide by $(\sec x+\tan x)$: $\int\frac{\sec^2x+\sec x\tan x}{\sec x+\tan x}\,dx$. Since $\frac{d}{dx}(\sec x+\tan x)=\sec x\tan x+\sec^2x$ exactly matches the numerator, this is $\log|\sec x+\tan x|+c$, which also equals $\log\left|\tan\left(\frac x2+\frac\pi4\right)\right|+c$.

$\int\csc x\,dx$. By the parallel trick, multiplying and dividing by $(\csc x-\cot x)$, $\int\csc x\,dx=\log|\csc x-\cot x|+c=\log\left|\tan\frac x2\right|+c$.

The same "numerator is the derivative of the denominator" idea, or "power of $f(x)$ times $f'(x)$", extends to many exponential, logarithmic and trigonometric integrands built from a hidden substitution:

$\int\dfrac{\cot(\log x)}{x}\,dx$. Put $\log x=t$, so $\frac1x\,dx=dt$; the integral becomes $\int\cot t\,dt=\log|\sin t|+c=\log|\sin(\log x)|+c$.

$\int\dfrac{\cos\sqrt x}{\sqrt x}\,dx$. Put $\sqrt x=t$, so $\frac{1}{2\sqrt x}\,dx=dt$, i.e. $\frac{1}{\sqrt x}\,dx=2\,dt$; the integral becomes $2\int\cos t\,dt=2\sin t+c=2\sin\sqrt x+c$.

$\int\dfrac{\sec^8x}{\csc x}\,dx$. Rewrite as $\sec^7x\cdot\sec x\cdot\frac{1}{\csc x}=\sec^7x\cdot\sec x\tan x$ (using $\frac1{\csc x}=\sin x=\cos x\tan x$, and grouping $\sec x\cdot\cos x\tan x=\tan x$, then $\sec^7x\cdot\tan x = \sec^6x\cdot\sec x\tan x$). Put $\sec x=t$, $\sec x\tan x\,dx=dt$: integral becomes $\int t^6\,dt=\frac{t^7}7+c=\frac{\sec^7x}7+c$.

$\int\dfrac{1}{x+\sqrt x}\,dx$. Factor the denominator as $\sqrt x(\sqrt x+1)$. Put $\sqrt x+1=t$, so $\frac{1}{2\sqrt x}dx=dt$, i.e. $\frac{1}{\sqrt x}dx=2\,dt$: integral becomes $2\int\frac1t\,dt=2\log|t|+c=2\log(\sqrt x+1)+c$.

$\int5^{5x}\cdot5^x\,dx$. Put $5^x=t$, so $5^x\log5\,dx=dt$; the integrand $5^{5x}\cdot5^x=t^5\cdot t=t^6$... more precisely rewriting via $t$: the integral becomes $\frac{1}{\log5}\int5^t\,dt=\frac{1}{\log5}\cdot\frac{5^t}{\log5}+c=\frac{1}{(\log5)^2}\cdot5^{5x}+c$.

$\int\dfrac{1}{1+e^{-x}}\,dx$. Multiply top and bottom by $e^x$: $\int\dfrac{e^x}{e^x+1}\,dx=\log(e^x+1)+c$, since the numerator is exactly the derivative of the denominator.

$\int e^x\cdot\dfrac{1+x}{\cos(xe^x)}\,dx$. Put $xe^x=t$; differentiating, $(xe^x+e^x)\,dx=e^x(1+x)\,dx=dt$. The integral becomes $\int\sec t\,dt=\log|\sec t+\tan t|+c=\log\left[\sec(xe^x)+\tan(xe^x)\right]+c$.

$\int\dfrac{1}{3^x+7x^{-n}}\,dx$. Rewrite the denominator as $\dfrac{3x^{n+1}+7}{x^n}$, so the integral is $\int\dfrac{x^n}{3x^{n+1}+7}\,dx$. Put $3x^{n+1}+7=t$, so $3(n+1)x^n\,dx=dt$: the integral becomes $\dfrac{1}{3(n+1)}\log|3x^{n+1}+7|+c$.

$\int(3x+2)\sqrt{x-4}\,dx$. Put $x-4=t$, so $x=4+t$, $dx=dt$: integral becomes $\int(14+3t)\sqrt t\,dt=\int\left(14t^{1/2}+3t^{3/2}\right)dt=\dfrac{28}{3}(x-4)^{3/2}+\dfrac{6}{5}(x-4)^{5/2}+c$.

$\int\dfrac{\sin(x+a)}{\cos(x-b)}\,dx$. Write $x+a=(x-b)+(a+b)$ and expand $\sin[(x-b)+(a+b)]$ by the addition formula, then divide by $\cos(x-b)$: the integrand becomes $\cos(a+b)\tan(x-b)+\sin(a+b)$, so the integral is $\cos(a+b)\log|\sec(x-b)|+x\sin(a+b)+c$.

$\int\dfrac{e^x+1}{e^x-1}\,dx$. Write the numerator as $(e^x-1)+2$: integral becomes $\int1\,dx+2\int\dfrac{1}{e^x-1}\,dx=x+2\int\dfrac{e^{-x}}{1-e^{-x}}\,dx$; putting $1-e^{-x}=t$ (so $e^{-x}\,dx=dt$) gives $x+2\log(1-e^{-x})+c$. …