求定积分：∫01xb−xaln⁡xdx\displaystyle\int_{0}^{1}\dfrac{x^b-x^a}{\ln x}\mathrm{d}x，其中 a,b>−1a, b>-1

令 I(t)=∫01xt−1ln⁡xdxI(t)=\displaystyle\int_{0}^{1}\dfrac{x^t-1}{\ln x}\mathrm{d}x，接下来考虑求该定积分，首先两边对 tt 求导

I′(t)=ddx∫01xt−1ln⁡xdx=∫01∂∂txt−1ln⁡xdx=x1+t1+t∣01=11+t\begin{align} I'(t) & =\dfrac{\mathrm{d}}{\mathrm{d}x}\int_{0}^{1}\dfrac{x^t-1}{\ln x}\mathrm{d}x\\ & = \int_{0}^{1}\dfrac{\partial}{\partial t}\dfrac{x^t-1}{\ln x}\mathrm{d}x\\ & = \left. \frac{x^{1+t}}{1+t}\right|_{0}^{1}\\ & = \dfrac{1}{1+t} \end{align}

则

I(t)=∫I′(t)dt=ln⁡∣1+t∣+CI(t)=\int I'(t)\mathrm{d}t=\ln|1+t|+C

接下来求 CC，因为 I(0)=0I(0)=0，所以 ln⁡1+C=0\ln 1+C=0，从而 C=0C=0，因此 I(t)=ln⁡∣t+1∣I(t)=\ln|t+1|，则

∫01xb−xaln⁡xdx=I(b)−I(a)=ln⁡∣1+b1+a∣=ln⁡b+1a+1\int_{0}^{1}\dfrac{x^b-x^a}{\ln x}\mathrm{d}x=I(b)-I(a)=\ln\left|\dfrac{1+b}{1+a}\right|=\ln\dfrac{b+1}{a+1}

求定积分：∫0π2ln⁡(a2sin⁡2x+b2cos⁡2x)dx\displaystyle\int_{0}^{\frac{\pi}{2}}\ln(a^2\sin^2x+b^2\cos^2x)\mathrm{d}x，其中 a,b>0a, b>0

为了巩固上一题的解法，我向 AI 要来了这题，这题的解法与上题有相似但不完全一样

令 I(a)=∫0π2ln⁡(a2sin⁡2x+b2cos⁡2x)dxI(a)=\displaystyle\int_{0}^{\frac{\pi}{2}}\ln(a^2\sin^2x+b^2\cos^2x)\mathrm{d}x，同样地，两边对 aa 求导

I′(a)=∫0π22asin⁡2xa2sin⁡2x+b2cos⁡2xdx=∫0π22atan⁡2xa2tan⁡2x+b2dxI'(a)=\int_{0}^{\frac{\pi}{2}}\dfrac{2a\sin^2 x}{a^2\sin^2x+b^2\cos^2 x}\mathrm{d}x=\int_{0}^{\frac{\pi}{2}}\frac{2a\tan^2 x}{a^2\tan^2 x+b^2}\mathrm{d}x

令 t=tan⁡xt=\tan x，则 dx=dt1+t2\mathrm{d}x=\dfrac{\mathrm{d}t}{1+t^2}

I′(a)=∫0+∞2at2(a2t2+b2)(1+t2)dt=−2ab2a2−b2∫0+∞1a2t2+b2dt+2aa2−b2∫0+∞11+t2dt=−2ab2a2−b21abarctan⁡atb∣0+∞+2aa2−b2arctan⁡t∣0+∞=π(a−b)a2−b2==πa+b\begin{align} I'(a) & =\int_{0}^{+\infty}\dfrac{2at^2}{(a^2t^2+b^2)(1+t^2)}\mathrm{d}t\\ & = -\frac{2ab^2}{a^2-b^2}\int_{0}^{+\infty}\frac{1}{a^2t^2+b^2}\mathrm{d}t+\frac{2a}{a^2-b^2}\int_{0}^{+\infty}\frac{1}{1+t^2}\mathrm{d}t\\ & = -\frac{2ab^2}{a^2-b^2}\left.\frac{1}{ab}\arctan\frac{at}{b}\right|_{0}^{+\infty}+\frac{2a}{a^2-b^2}\left.\arctan t\right|_0^{+\infty}\\ & = \frac{\pi(a-b)}{a^2-b^2}\\= & = \frac{\pi}{a+b} \end{align}

则

I(a)=∫πa+bda=πln⁡∣a+b∣+CI(a)=\int\frac{\pi}{a+b}\mathrm{d}a=\pi\ln|a+b|+C

又 I(b)=∫0π2ln⁡(b2)dx=πln⁡bI(b)=\displaystyle\int_{0}^{\frac{\pi}{2}}\ln(b^2)\mathrm{d}x=\pi\ln b，则 C=πln⁡b−πln⁡∣2b∣=−πln⁡2C=\pi \ln b-\pi\ln|2b|=-\pi\ln 2

所以

∫0π2ln⁡(a2sin⁡2x+b2cos⁡2x)dx=πln⁡a+b2\int_{0}^{\frac{\pi}{2}}\ln(a^2\sin^2x+b^2\cos^2x)\mathrm{d}x=\pi\ln\dfrac{a+b}{2}

求定积分：∫xsin⁡3xdx\displaystyle\int x\sin^3x\mathrm{d}x

求不定积分：∫sec⁡3xdx\displaystyle\int\sec^3x\mathrm{d}x

首先凑微分 ∫sec⁡xd(tan⁡x)\displaystyle\int\sec x\mathrm{d}(\tan x)，接下来用分部积分公式

∫sec⁡3xdx=∫sec⁡xd(tan⁡x)=sec⁡xtan⁡x−∫tan⁡2xsec⁡xdx=sec⁡xtan⁡x−∫(sec⁡2x−1)sec⁡xdx=sec⁡xtan⁡x−∫sec⁡3xdx+∫sec⁡xdx\begin{align} & \int\sec^3 x\mathrm{d}x\\ = & \int\sec x\mathrm{d}(\tan x)\\ = & \sec x\tan x-\int\tan^2 x\sec x\mathrm{d}x\\ = & \sec x\tan x-\int(\sec^2x -1)\sec x\mathrm{d}x\\ = & \sec x\tan x-\int\sec^3x\mathrm{d}x+\int\sec x\mathrm{d}x \end{align}

则有

∫sec⁡3xdx=12(sec⁡xtan⁡x+∫sec⁡xdx)\int\sec^3 x\mathrm{d}x=\frac{1}{2}\left(\sec x\tan x+\int\sec x\mathrm{d}x\right)

那么只要求出 ∫sec⁡xdx\displaystyle\int\sec x\mathrm{d}x 求行了，这里我们设 u=sec⁡x+tan⁡xu=\sec x+\tan x，则

du=sec⁡x(sec⁡x+tan⁡x)dx=usec⁡xdx\mathrm{d}u=\sec x(\sec x+\tan x)\mathrm{d}x=u\sec x\mathrm{d}x

则 sec⁡xdx=duu\sec x\mathrm{d}x=\dfrac{\mathrm{d}u}{u}，则

∫sec⁡xdx=∫duu=ln⁡∣u∣=ln⁡∣sec⁡x+tan⁡x∣+C\int\sec x\mathrm{d}x=\int\frac{\mathrm{d}u}{u}=\ln|u|=\ln|\sec x+\tan x|+C

代回即可得到

∫sec⁡xdx=12(sec⁡xtan⁡x+ln⁡∣sec⁡x+tan⁡x∣)+C\int\sec x\mathrm{d}x=\frac{1}{2}\left(\sec x\tan x+\ln|\sec x+\tan x|\right)+C

求不定积分：∫sin⁡2nxsin⁡xdx\displaystyle\int\frac{\sin 2nx}{\sin x}\mathrm{d}x

注意到

sin⁡2nx=∑k=0n[sin⁡2kx−sin⁡(2k−2)x]=2sin⁡x∑k=0ncos⁡(2k−2)x\begin{align} \sin 2nx & =\sum_{k=0}^{n}[\sin2kx-\sin(2k-2)x]\\ & = 2\sin x\sum_{k=0}^{n}\cos(2k-2)x \end{align}

则

∫sin⁡2nxsin⁡xdx=∫2∑k=0ncos⁡(2k−2)xdx=2∑k=0n∫cos⁡(2k−2)xdx=2∑k=0nsin⁡(2k−1)x2k−1+C\begin{align} \int\frac{\sin 2nx}{\sin x}\mathrm{d}x & =\int2\sum_{k=0}^{n}\cos(2k-2)x\mathrm{d}x\\ & =2\sum_{k=0}^{n}\int\cos(2k-2)x\mathrm{d}x\\ & =2\sum_{k=0}^{n}\frac{\sin(2k-1)x}{2k-1}+C \end{align}

同样的，我们还可以通过这种方法求得 ∫sin⁡(2n+1)xsin⁡xdx=x+2∑k=1nsin⁡2kx2k+C\displaystyle\int\dfrac{\sin(2n+1)x}{\sin x}\mathrm{d}x=x+2\sum_{k=1}^{n}\dfrac{\sin2kx}{2k}+C

求定积分：∫0π1−cos⁡(nx)1−cos⁡xdx\displaystyle\int_{0}^{\pi}\dfrac{1-\cos(nx)}{1-\cos x}\mathrm{d}x，其中 n∈Nn\in\N

首先利用半角公式将形式简化

∫0π1−cos⁡(nx)1−cos⁡xdx=∫0πsin⁡2nx2sin⁡2x2dx=x=2u2∫0π2sin⁡2nusin⁡2udu\int_{0}^{\pi}\dfrac{1-\cos(nx)}{1-\cos x}\mathrm{d}x=\int_{0}^{\pi}\frac{\sin^2\frac{nx}{2}}{\sin^2\frac{x}{2}}\mathrm{d}x\xlongequal{x=2u}2\int_{0}^{\frac{\pi}{2}}\frac{\sin^2nu}{\sin^2u}\mathrm{d}u

令 I(n)=∫0π2sin⁡2nusin⁡2udxI(n)=\displaystyle\int_{0}^{\frac{\pi}{2}}\frac{\sin^2nu}{\sin^2u}\mathrm{d}x，利用三角平方差公式可得

I(n)−I(n−1)=∫0π2sin⁡usin⁡[(2n−1)u]sin⁡2udu=∫0π2sin⁡[(2n−1)u]sin⁡uduI(n)-I(n-1)=\int_{0}^{\frac{\pi}{2}}\frac{\sin u\sin\left[(2n-1)u\right]}{\sin^2 u}\mathrm{d}u=\int_{0}^{\frac{\pi}{2}}\frac{\sin\left[(2n-1)u\right]}{\sin u}\mathrm{d}u

想要消去分母的 sin⁡u\sin u，我们需要再做差

[I(n+1)−I(n)]−[I(n)−I(n−1)]=∫0π2sin⁡[(2n+1)u]−sin⁡[(2n−1)u]sin⁡udx=∫0π22cos⁡2nusin⁡usin⁡udu=2∫0π2cos⁡2nudu=0\begin{align} [I(n+1)-I(n)]-[I(n)-I(n-1)] & =\int_{0}^{\frac{\pi}{2}}\frac{\sin\left[(2n+1)u\right]-\sin\left[(2n-1)u\right]}{\sin u}\mathrm{d}x\\ & =\int_{0}^{\frac{\pi}{2}}\frac{2\cos 2nu\sin u}{\sin u}\mathrm{d}u\\ & = 2\int_{0}^{\frac{\pi}{2}}\cos 2nu\mathrm{d}u\\ & = 0 \end{align}

则 I(n)−I(n−1)=I(1)−I(0)=π2I(n)-I(n-1)=I(1)-I(0)=\dfrac{\pi}{2}，则根据等差数列的通项公式有

I(n)=π2nI(n)=\dfrac{\pi}{2}n

则

∫0π1−cos⁡(nx)1−cos⁡xdx=2I(n)=nπ\boxed{\int_{0}^{\pi}\dfrac{1-\cos(nx)}{1-\cos x}\mathrm{d}x=2I(n)=n\pi}

实际上，此题和上一题是同一类题，因此我们也可以采用上一题的解法

首先仍然采用半角公式和换元，此处省略， 直接计算 ∫0π2sin⁡2nusin⁡2udx\displaystyle\int_{0}^{\frac{\pi}{2}}\frac{\sin^2nu}{\sin^2u}\mathrm{d}x

∫0π2sin⁡2nusin⁡2udu=∫0π2∑i=1n[sin⁡2(iu)−sin⁡2(i−1)u]sin⁡2udu=∫0π2sin⁡u∑i=1nsin⁡(2i−1)usin⁡2udu=∫0π2∑i=1n{∑j=1i[sin⁡(2j−1)u−sin⁡(2j−3)u]+sin⁡u}sin⁡udu=∫0π2∑i=1n{2∑j=1i[cos⁡(2j−2)sin⁡u]+sin⁡u}sin⁡udu=∫0π2[2∑i=1n∑j=1icos⁡(2j−2)u+n]du=2∑i=1n∑j=1i∫0π2cos⁡(2j−2)udu+nπ2=nπ2\begin{align} \displaystyle\int_{0}^{\frac{\pi}{2}}\frac{\sin^2nu}{\sin^2u}\mathrm{d}u & = \int_{0}^{\frac{\pi}{2}}\frac{\sum_{i=1}^{n}[\sin^2(iu)-\sin^2(i-1)u]}{\sin^2 u}\mathrm{d}u\\ & =\int_{0}^{\frac{\pi}{2}}\frac{\sin u\sum_{i=1}^{n}\sin(2i-1)u}{\sin ^2u}\mathrm{d}u\\ & =\int_{0}^{\frac{\pi}{2}}\frac{\sum_{i=1}^{n}\left\{\sum_{j=1}^{i}\left[\sin(2j-1)u-\sin(2j-3)u\right]+\sin u\right\}}{\sin u}\mathrm{d}u\\ & =\int_{0}^{\frac{\pi}{2}}\frac{\sum_{i=1}^{n}\left\{2\sum_{j=1}^{i}\left[\cos(2j-2)\sin u\right]+\sin u\right\}}{\sin u}\mathrm{d}u\\ & =\int_{0}^{\frac{\pi}{2}}\left[2\sum_{i=1}^{n}\sum_{j=1}^{i}\cos(2j-2)u+n\right]\mathrm{d}u\\ & =2\sum_{i=1}^{n}\sum_{j=1}^{i}\int_{0}^{\frac{\pi}{2}}\cos(2j-2)u\mathrm{d}u+\dfrac{n\pi}{2}\\ & =\dfrac{n\pi}{2} \end{align}

因此

∫0π1−cos⁡(nx)1−cos⁡xdx=nπ\boxed{\int_{0}^{\pi}\dfrac{1-\cos(nx)}{1-\cos x}\mathrm{d}x=n\pi}

求不定积分： ∫eaxcos⁡bxdx\displaystyle\int e^{ax}\cos bx\mathrm{d}x，∫eaxsin⁡bxdx\displaystyle\int e^{ax}\sin bx\mathrm{d}x

首先对两个积分分别使用分部积分法

∫eaxcos⁡bxdx=1aeaxcos⁡bx+ba∫eaxsin⁡bxdx∫eaxsin⁡bxdx=1aeaxsin⁡bx−ba∫eaxcos⁡bxdx\begin{align} & \int e^{ax}\cos bx\mathrm{d}x=\frac{1}{a}e^{ax}\cos bx+\frac{b}{a}\int e^{ax}\sin bx\mathrm{d}x\\ & \int e^{ax}\sin bx\mathrm{d}x=\frac{1}{a}e^{ax}\sin bx-\frac{b}{a}\int e^{ax}\cos bx\mathrm{d}x \end{align}

这里发现，我们得到了关于这两个积分的二元一次方程组，解这个方程组可得

∫eaxcos⁡bxdx=eax(acos⁡bx+bsin⁡bx)a2+b2+C∫eaxsin⁡bxdx=eax(asin⁡bx−bcos⁡bx)a2+b2+C\boxed{ \begin{align} & \int e^{ax}\cos bx\mathrm{d}x=\frac{e^{ax}(a\cos bx+b\sin bx)}{a^2+b^2}+C\\ & \int e^{ax}\sin bx\mathrm{d}x=\frac{e^{ax}(a\sin bx-b\cos bx)}{a^2+b^2}+C \end{align} }

求不定积分：∫cos⁡2xsin⁡x+cos⁡xdx\displaystyle\int\dfrac{\cos^2 x}{\sin x+\cos x}\mathrm{d}x

先来看第一种解法，这种解法的关键是要想到使用半角公式

∫cos⁡2xsin⁡x+cos⁡x dx=12∫1+cos⁡2xsin⁡x+cos⁡x dx=12∫1+(cos⁡x+sin⁡x)(cos⁡x−sin⁡x)sin⁡x+cos⁡x dx=12∫(1sin⁡x+cos⁡x+cos⁡x−sin⁡x) dx=12[∫1sin⁡x+cos⁡x dx+∫(cos⁡x−sin⁡x) dx]=12[∫12sin⁡(x+π4) dx+sin⁡x+cos⁡x]=12[12∫csc⁡(x+π4) dx+sin⁡x+cos⁡x]=12[12ln⁡∣csc⁡(x+π4)−cot⁡(x+π4)∣+sin⁡x+cos⁡x]+C\begin{align*} \int \frac{\cos^2 x}{\sin x + \cos x} \,\mathrm{d}x &= \frac{1}{2} \int \frac{1 + \cos 2x}{\sin x + \cos x} \,\mathrm{d}x \\ &= \frac{1}{2} \int \frac{1 + (\cos x + \sin x)(\cos x - \sin x)}{\sin x + \cos x} \,\mathrm{d}x \\ &= \frac{1}{2} \int \left( \frac{1}{\sin x + \cos x} + \cos x - \sin x \right) \,\mathrm{d}x \\ &= \frac{1}{2} \left[ \int \frac{1}{\sin x + \cos x} \,\mathrm{d}x + \int (\cos x - \sin x) \,\mathrm{d}x \right] \\ &= \frac{1}{2} \left[ \int \frac{1}{\sqrt{2} \sin\left(x + \frac{\pi}{4}\right)} \,\mathrm{d}x + \sin x + \cos x \right] \\ &= \frac{1}{2} \left[ \frac{1}{\sqrt{2}} \int \csc\left(x + \frac{\pi}{4}\right) \,\mathrm{d}x + \sin x + \cos x \right] \\ &= \frac{1}{2} \left[ \frac{1}{\sqrt{2}} \ln \left| \csc\left(x + \frac{\pi}{4}\right) - \cot\left(x + \frac{\pi}{4}\right) \right| + \sin x + \cos x \right] + C \end{align*}

第二种解法与上一题有些相似之处（利用对偶式）

令 I=∫cos⁡2xsin⁡x+cos⁡xdxI=\displaystyle\int\dfrac{\cos^2 x}{\sin x+\cos x}\mathrm{d}x，J=∫sin⁡2xsin⁡x+cos⁡xdxJ=\displaystyle\int\dfrac{\sin^2 x}{\sin x+\cos x}\mathrm{d}x

然后分别计算 I+JI+J 和 I−JI-J

I+J=∫dxsin⁡x+cos⁡x=12ln⁡∣csc⁡(x+π4)−cot⁡(x+π4)∣+C1I−J=∫(cos⁡x−sin⁡x)dx=sin⁡x+cos⁡x+C2\begin{align} & I+J = \int\frac{\mathrm{d}x}{\sin x+\cos x}=\frac{1}{\sqrt{2}}\ln\left|\csc\left(x+\frac{\pi}{4}\right)-\cot\left(x+\frac{\pi}{4}\right)\right|+C_1\\ & I-J = \int(\cos x-\sin x)\mathrm{d}x=\sin x+\cos x+C_2 \end{align}

两式相加即可得到 II