# 三角换元法

$\int_M \mathrm{d}\omega = \oint_{\partial M} \omega$

## 含有a2 − x2的积分

$\int\frac{dx}{\sqrt{a^2-x^2}}$

$x=a\sin(\theta),\ dx=a\cos(\theta)\,d\theta$
$\theta=\arcsin\left(\frac{x}{a}\right)$

$\int\frac{dx}{\sqrt{a^2-x^2}} = \int\frac{a\cos(\theta)\,d\theta}{\sqrt{a^2-a^2\sin^2(\theta)}} = \int\frac{a\cos(\theta)\,d\theta}{\sqrt{a^2(1-\sin^2(\theta))}}$
${} = \int\frac{a\cos(\theta)\,d\theta}{\sqrt{a^2\cos^2(\theta)}} = \int d\theta=\theta+C=\arcsin\left(\frac{x}{a}\right)+C$

$\int_0^{a/2}\frac{dx}{\sqrt{a^2-x^2}} =\int_0^{\pi/6}d\theta=\frac{\pi}{6}.$

## 含有a2 + x2的积分

$\int\frac{dx}{a^2+x^2}$

$x=a\tan(\theta),\ dx=a\sec^2(\theta)\,d\theta$
$\theta=\arctan\left(\frac{x}{a}\right)$

\begin{align} & {} \quad \int\frac{dx}{a^2+x^2} = \int\frac{a\sec^2(\theta)\,d\theta}{a^2+a^2\tan^2(\theta)} = \int\frac{a\sec^2(\theta)\,d\theta}{a^2(1+\tan^2(\theta))} \\ & {} = \int \frac{a\sec^2(\theta)\,d\theta}{a^2\sec^2(\theta)} = \int \frac{d\theta}{a} = \frac{\theta}{a}+C = \frac{1}{a} \arctan \left(\frac{x}{a}\right)+C \end{align}

a > 0）。

## 含有x2 − a2的积分

$\int\frac{dx}{x^2 - a^2}$

$\int\sqrt{x^2 - a^2}\,dx$

$x = a \sec(\theta),\ dx = a \sec(\theta)\tan(\theta)\,d\theta$
$\theta = \arcsec\left(\frac{x}{a}\right)$
\begin{align} & {} \quad \int\sqrt{x^2 - a^2}\,dx = \int\sqrt{a^2 \sec^2(\theta) - a^2} \cdot a \sec(\theta)\tan(\theta)\,d\theta \\ & {} = \int\sqrt{a^2 (\sec^2(\theta) - 1)} \cdot a \sec(\theta)\tan(\theta)\,d\theta = \int\sqrt{a^2 \tan^2(\theta)} \cdot a \sec(\theta)\tan(\theta)\,d\theta \\ & {} = \int a^2 \sec(\theta)\tan^2(\theta)\,d\theta = a^2 \int \sec(\theta)\ (\sec^2(\theta) - 1)\,d\theta \\ & {} = a^2 \int (\sec^3(\theta) - \sec(\theta))\,d\theta. \end{align}

## 含有三角函数的积分

$\int f(\sin x,\cos x)\,dx=\int\frac1{\pm\sqrt{1-u^2}}f\left(u,\pm\sqrt{1-u^2}\right)\,du, \qquad \qquad u=\sin x$
$\int f(\sin x,\cos x)\,dx=\int\frac{-1}{\pm\sqrt{1-u^2}}f\left(\pm\sqrt{1-u^2},u\right)\,du \qquad \qquad u=\cos x$
$\int f(\sin x,\cos x)\,dx=\int\frac2{1+u^2} f\left(\frac{2u}{1+u^2},\frac{1-u^2}{1+u^2}\right)\,du \qquad \qquad u=\tan\frac x2$
$\int\frac{\cos x}{(1+\cos x)^3}\,dx = \int\frac2{1+u^2}\frac{\frac{1-u^2}{1+u^2}}{\left(1+\frac{1-u^2}{1+u^2}\right)^3}\,du =$
$\frac{1}{4}\int(1-u^4)\,du = \frac{1}{4}\left(u-\frac15u^5\right) + C = \frac{(1+3\cos x+\cos^2x)\sin x}{5(1+\cos x)^3} + C$