The integration task

(1)

(2)

(3)

(4)

Proof of the formulae (4): Using the double angle formulas of sine and cosine and then dividing the numerators and the denominators by $\cos^2\frac{x}{2}$ we obtain $$\sin{x} = \frac{2\sin\frac{x}{2}\cos\frac{x}{2}}{\sin^2\frac{x}{2}+\cos^2\frac{x}{2}} = \frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}} = \frac{2t}{1+t^2},$$ $$\cos{x} = \frac{\cos^2\frac{x}{2}-\sin^2\frac{x}{2}}{\sin^2\frac{x}{2}+\cos^2\frac{x}{2}} = \frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}} = \frac{1-t^2}{1+t^2}.$$

Example. The above formulae give from $\displaystyle \int\frac{dx}{\sin{x}}$ the result $$\int\frac{dx}{\sin{x}} = \int\frac{1\!+\!t^2}{2t}\cdot2\cdot\frac{1}{1\!+\!t^2}\;dt = \int\frac{dt}{t} = \ln|t|+C = \ln\left|\tan\frac{x}{2}\right|+C$$ (which can also be expressed in the form $-\ln|\csc{x}+\cot{x}|+C$ ; see the goniometric formulas).

Note. The substitution (2) is sometimes called the ``universal trigonometric substitution''. In practice, it often gives rational functions that are too complicated. In many cases, it is more profitable to use other substitutions:

• In the case $\int\!R(\sin{x})\cos{x}\,dx$ the substitution $\sin{x} = t$ is simpler.
• Similarly, in the case $\int\!R(\cos{x})\sin{x}\,dx$ the substitution $\cos{x} = t$ is simpler.
• If the integrand depends only on $\tan{x}$ , the substitution $\tan{x} = t$ is simpler.
• If the integrand is of the form $R(\sin^2{x},\, \cos^2{x})$ , one can use the substitution $\tan{x} = t$ ; then
  $\displaystyle \cos^2{x} = \frac{1}{1+\tan^2{x}} = \frac{1}{1+t^2}$ , $\displaystyle \sin^2{x} = 1-\cos^2{x} = \frac{t^2}{1+t^2}$ , $\displaystyle dx = \frac{dt}{1+t^2}.$

Example. The integration of $\displaystyle \int\!\frac{dx}{\cos^4{x}}\,dx$ is of the last case: $$\int\!\frac{dx}{\cos^4{x}}\,dx = \int\!\frac{1}{(\cos^2{x})^2}\,dx = \int\!(1+t^2)^2\cdot\frac{dt}{1+t^2} = \int\!(1+t^2)\,dt = \frac{t^3}{3}+t+C = \frac{1}{3}\tan^3{x}+\tan{x}+C.$$

Example. The integral $\displaystyle I = \int\!\frac{dx}{\cos^3{x}}\,dx = \int\! \sec^3{x}\,dx $ is a peculiar case in which one does not have to use the substitutions mentioned above, as integration by parts is a simpler method for evaluating this integral. Thus,

$$u = \sec{x}\; \Rightarrow\; du = \sec{x}\;\tan{x}\,dx; \qquad dv = \sec^2{x}\,dx \; \Rightarrow \; v = \tan{x}.$$ Therefore,