Додаток: інтегрування з використанням формули половинного кута ($u=\tan(x/2)$)

The Tangent Half-Angle Substitution

The substitution $u = \tan{\frac{x}{2}}$ has been called by Spivak "the world's sneakiest substitution". Suppose your goal is to integrate

$$\int{\frac{dx}{3-5\sin{x}}}.$$

What would you do? There are four main basic methods of integration - partial fractions, parts, trigonometric substitution, and $u-$ substitution. The first doesn't work because the denominator is already a binomial that can't be factored, and the second because there isn't a way to find two functions that would simplify the problem. Trigonometric substitution fails because if you substitute an inverse trig function, an ugly differential appears, and if you substitute a regular trig function, it becomes even worse as you wind up with nested trigonometric functions. Thus $u-$ substitution is the only remaining tool that we have left. But finding a good substitution is difficult, as substituting a regular function of $x$ or $2x$ doesn't help because you have the $dx$ term still unaccounted for. This is where the $u = \tan{\frac{x}{2}}$ substitution comes in.

Before applying this substitution, we should probably investigate exactly how strong it truly is. Firstly, from the basic double-angle formula for sine, $\sin{x} = 2\sin{\frac{x}{2}}\cos{\frac{x}{2}} = 2\tan{\frac{x}{2}}\cos^2{\frac{x}{2}}$. Now by a bit of manipulation we can express the result entirely in terms of our new $u$: we have that $2\tan{\frac{x}{2}}\cos^2{\frac{x}{2}} = \frac{2u}{\sec^2{\frac{x}{2}}} = \frac{2u}{1 + \tan^2{\frac{u}{2}}} = \frac{2u}{1+u^2}$.

Similarly, we can express $\cos{x}$ in terms of $u$ as well: we have that $\cos{x} = 1 - 2\sin^2{\frac{x}{2}} = 1 - 2\tan^2{\frac{x}{2}}\cos^2{\frac{x}{2}} \\= 1 - 2u^2 (\frac{1}{1+u^2})=\frac{1+u^2-2u^2}{1+u^2}=\frac{1-u^2}{1+u^2}.$

Finally, we can also find $dx$ in terms of $u$ and $du$, since $\frac{du}{dx}=\frac{1}{2}\sec^2{\frac{x}{2}} = \frac{1 + \tan^2{\frac{x}{2}}}{2} = \frac{1 + u^2}{2}$, so thus $dx = \frac{2du}{1+u^2}$.

In our particular case, we have

$$\int{\frac{dx}{3-5\sin{x}}} = \int{\frac{\frac{2du}{1+u^2}}{3-5(\frac{2u}{1+u^2})}}$$

This still looks a bit messy, but now multiplying the numerator and denominator by $1 + u^2$ and rearranging a bit gives

$$\int{\frac{\frac{2du}{1+u^2}}{3-5(\frac{2u}{1+u^2})}} = \int{\frac{2du}{3({1+u^2})-5(2u)}}=\int{\frac{2du}{3u^2 - 10u + 3}},$$

which is a lot easier to evaluate than our original integral. In fact, the denominator factors nicely to $(3u-1)(u-3)$, and additionally we have that $2 = \frac{1}{4}(3u-1) - \frac{3}{4}(u-3)$ Thus we have that

$$\int{\frac{2du}{3u^2 - 10u + 3}} = \int{du\bigg(\frac{1}{4(u-3)}-\frac{3}{4(3u-1)}\bigg)}.$$

Now the rest is easy.

$$\int{du\bigg(\frac{1}{4(u-3)}- \frac{3}{4(3u-1)}\bigg)} \\= \frac{1}{4}(\ln{|u-3|} - \ln{|3u-1|}) + C = \frac{1}{4}\ln{\Bigg|\frac{u-3}{3u-1}\Bigg|}+C.$$

Finally, resubstitute back $\tan{\frac{x}{2}}$ for $u$:

$$\frac{1}{4}\ln{\Bigg|\frac{u-3}{3u-1}\Bigg|}+C=\frac{1}{4}\ln{\Bigg|\frac{\tan{\frac{x}{2}}-3}{3\tan{\frac{x}{2}}-1}\Bigg|}+C$$

Thus we have

$$\int{\frac{dx}{3-5\sin{x}}} = \frac{1}{4}\ln{\Bigg|\frac{\tan{\frac{x}{2}}-3}{3\tan{\frac{x}{2}}-1}\Bigg|}+C.$$

While the result isn't a nice expression, at least we were able to antidifferentiate the given function. In fact, this method can reduce any rational function in terms of sine and cosine to a different rational function in terms of just the new variable $u$.