Week11 Homework1
- 4.2.1(1)
\[\int\frac{1}{x^2+x-2}dx=\int\frac{1}{(x+2)(x-1)}dx=\int\frac{1}{3(x-1)}dx-\int\frac{1}{3(x+2)}dx=\frac{1}{3}(\ln|x-1|+\ln|x+2|)+C
\]
- 4.2.1(2)
\[\int\frac{x^4}{x^2+1}dx=\int(x^2-1+\frac{1}{x^2+1})dx=\frac{x^3}{3}-x+\arctan x+C
\]
- 4.2.1(3)
\[\int\frac{x^3+1}{x^3-x}dx=\int\frac{x^2-x+1}{x^2-x}dx=\int(1+\frac{1}{x(x-1)})dx=\int(1+\frac{1}{x-1}-\frac{1}{x})dx=x+\ln|x-1|+\ln|x|+C
\]
- 4.2.1(4)
\[\int\frac{1}{(x^2+1)(x^2+x)}dx=\int\frac{1}{x(x+1)(x^2+1)}dx=\int(\frac{1}{x}-\frac{1}{2(x+1)}-\frac{2x}{4(x^2+1)}-\frac{1}{2(x^2+1)})dx=\ln|x|-\frac{1}{2}\ln|x+1|-\frac{1}{4}\ln(x^2+1)-\frac{1}{2}\arctan x
\]
- 4.2.1(5)
\[\int\frac{x}{(x+1)^2(x^2+x+1)}dx=\int(\frac{1}{x^2+x+1}-\frac{1}{x^2+2x+1})dx=\int\frac{1}{x^2+x+1}dx+\frac{1}{x+1}
\]
\[\int\frac{1}{x^2+x+1}dx=\int\frac{1}{(x+\frac{1}{2})^2+\frac{3}{4}}dx
\]
设 \(x=t-\frac{1}{2},dx=dt\)
\[\int\frac{1}{x^2+x+1}=\int\frac{1}{t^2+\frac{3}{4}}dt
\]
设 \(t=\frac{\sqrt{3}}{2}s,dt=\frac{\sqrt{3}}{2}ds\)
\[\int\frac{1}{t^2+\frac{3}{4}}dt=\frac{2\sqrt{3}}{3}\int\frac{1}{s^2+1}ds=\frac{2\sqrt{3}}{3}\arctan s=\frac{2\sqrt{3}}{3}\arctan\frac{2\sqrt{3}}{3}t
\]
\[\int\frac{x}{(x+1)^2(x^2+x+1)}dx=\arctan\frac{2\sqrt{3}}{3}(x+\frac{1}{2})+\frac{1}{x+1}+C
\]
- 4.2.1(6)
\[\int\frac{x^2+1}{x^4+1}dx=\int\frac{1}{2}(\frac{1}{x^2-\sqrt{2}x+1}+\frac{1}{x^2+\sqrt{2}x+1})dx
\]
\[\int\frac{1}{x^2-\sqrt{2}x+1}dx=\int\frac{1}{(x-\frac{\sqrt{2}}{2})^2+\frac{1}{2}}dx
\]
设 \(x=t+\frac{\sqrt{2}}{2},dx=dt\)
\[\int\frac{1}{x^2-\sqrt{2}x+1}dx=\int\frac{1}{t^2+\frac{1}{2}}dt
\]
设 \(t=\frac{\sqrt{2}}{2}s,dt=\frac{\sqrt{2}}{2}ds\)
\[\int\frac{1}{t^2+\frac{1}{2}}dt=\sqrt{2}\int\frac{1}{s^2+1}ds=\sqrt{2}\arctan s=\sqrt{2}\arctan\sqrt{2}t
\]
\[\int\frac{1}{x^2-\sqrt{2}x+1}dx=\sqrt{2}\arctan\sqrt{2}(x-\frac{\sqrt{2}}{2})
\]
\[\int\frac{1}{x^2+\sqrt{2}x+1}dx=\int\frac{1}{(x+\frac{\sqrt{2}}{2})^2+\frac{1}{2}}dx
\]
设 \(x=t-\frac{\sqrt{2}}{2},dx=dt\)
\[\int\frac{1}{x^2+\sqrt{2}x+1}dx=\int\frac{1}{t^2+\frac{1}{2}}dt
\]
设 \(t=\frac{\sqrt{2}}{2}s,dt=\frac{\sqrt{2}}{2}ds\)
\[\int\frac{1}{t^2+\frac{1}{2}}dt=\sqrt{2}\int\frac{1}{s^2+1}ds=\sqrt{2}\arctan s=\sqrt{2}\arctan\sqrt{2}t
\]
\[\int\frac{1}{x^2+\sqrt{2}x+1}dx=\sqrt{2}\arctan\sqrt{2}(x+\frac{\sqrt{2}}{2})
\]
\[\int\frac{x^2+1}{x^4+1}dx=\frac{\sqrt{2}}{2}(\arctan\sqrt{2}(x-\frac{\sqrt{2}}{2})+\arctan\sqrt{2}(x+\frac{\sqrt{2}}{2}))
\]
- 习题 4.2.1(7)
\[\int\frac{x^5-x}{x^8+1}dx=
\]
- 习题 4.2.1(8)
\[\int\frac{x^{15}}{(x^8+1)^2}dx=\int(\frac{x^{15}+x^7}{x^{16}+2x^8+1}-\frac{x^7}{x^{16}+2x^8+1})dx=\frac{1}{16}\ln|(x^8+1)^2|+\frac{1}{8(x^8+1)}+C
\]
- 习题 4.2.2(1)
\[\int\frac{1+\sin x}{\sin x(1+\cos x)}dx
\]
设 \(t=\tan\frac{x}{2},dx=\frac{2}{1+t^2}dt\)
\[\int\frac{1+\sin x}{\sin x(1+\cos x)}dx=\int\frac{(t+1)^2(t^2+1)}{4t}\times\frac{2}{1+t^2}dt=\int(\frac{1}{2}t+1+\frac{1}{2t})dt=\frac{1}{4}t^2+t+\frac{1}{2}\ln|t|+C
\]
\[\int\frac{1+\sin x}{\sin x(1+\cos x)}dx=\frac{1}{4}\tan^2\frac{x}{2}+\tan\frac{x}{2}+\frac{1}{2}\ln|\tan\frac{x}{2}|+C
\]
- 习题 4.2.2(2)
\[\int\frac{\sin^5x}{\cos x}dx=\int\frac{\sin^5x}{\cos x(\sin^4x+2\sin^2x\cos^2x+\cos^4x)}dx
\]
设 \(t=\tan x,dx=\frac{1}{1+t^2}dt\)
\[\int\frac{\sin^5x}{\cos x}dx=\int\frac{t^5}{t^4+2t^2+1}\times\frac{1}{1+t^2}dt=\int\frac{t^5}{(t^2+1)^3}dt=\int(\frac{6t^5+12t^3+6t}{6(t^2+1)^3}-\frac{2t}{2(t^2+1)^3}-\frac{2t^3}{(t^2+1)^3})dx
\]
\[\int\frac{\sin^5x}{\cos x}dx=\frac{1}{2}\ln(t^2+1)+\frac{1}{2(t^2+1)}-2\int\frac{t^3}{(t^2+1)^3}dt
\]
设 \(t=\sinh\theta,dt=\cosh\theta d\theta\)
\[\int\frac{t^3}{(t^2+1)^3}dt=\int\frac{\tanh^3\theta}{\cosh^2\theta}d\theta=\frac{1}{4}\tanh^4\theta+C_1=\frac{1}{4}(\frac{t+\sqrt{t^2+1}-\frac{1}{t+\sqrt{t^2+1}}}{t+\sqrt{t^2+1}+\frac{1}{t+\sqrt{t^2+1}}})^4+C_1
\]
\[\int\frac{\sin^5x}{\cos x}dx=\frac{1}{2}\ln(\tan^2x+1)+\frac{1}{2(\tan^2x+1)}-\frac{1}{2}(\frac{\tan x+\sqrt{\tan^2x+1}-\frac{1}{\tan x+\sqrt{\tan^2x+1}}}{\tan x+\sqrt{\tan^2x+1}+\frac{1}{\tan x+\sqrt{\tan^2x+1}}})^4+C
\]
- 习题 4.2.2(3)
\[\int\frac{1}{\sin^4x\cos^2x}dx=\int\frac{(\sin^2x+\cos^2x)^3}{\sin^4x\cos^2x}dx
\]
设 \(t=\tan x,dx=\frac{dt}{1+t^2}\)
\[\int\frac{1}{\sin^4x\cos^2x}dx=\int\frac{t^4+2t^2+1}{t^4}dt=\int(1+\frac{2}{t^2}+\frac{1}{t^4})dt=t-\frac{2}{t}-\frac{1}{3t^3}+C=\tan x-\frac{2}{\tan x}-\frac{1}{3\tan^3x}+C
\]
-
习题 4.2.2(4)
-
习题 4.2.2(5)
\[\int\frac{\sin x\cos x}{1+\sin^4x}dx
\]
设 \(t=\tan x,dx=\frac{dt}{1+t^2}\)
\[\int\frac{\sin x\cos x}{1+\sin^4x}dx=\int\frac{t}{2t^4+2t^2+1}dt=\int\frac{t}{2[(t^2+\frac{1}{2})^2+\frac{1}{4}]}dt
\]
设 \(s=t^2+\frac{1}{2},ds=2tdt\)
\[\int\frac{\sin x\cos x}{1+\sin^4x}dx=\int\frac{1}{4(s^2+\frac{1}{4})}ds
\]
设 \(s=\frac{1}{2}u,ds=\frac{1}{2}du\)
\[\int\frac{\sin x\cos x}{1+\sin^4x}dx=\frac{1}{2}\int\frac{1}{u^2+1}du=\frac{1}{2}\arctan u+C=\frac{1}{2}\arctan2s+C=\frac{1}{2}\arctan(2t^2+1)+C=\frac{1}{2}\arctan(2\tan^2x+1)+C
\]
- 习题 4.2.2(6)
\[\int\frac{\sin^2x}{1+\sin^2x}dx
\]
设 \(t=\tan x,dx=\frac{dt}{1+t^2}\)
\[\int\frac{\sin^2x}{1+\sin^2x}dx=\int\frac{t^2}{(2t^2+1)(t^2+1)}dt=\int(\frac{1}{t^2+1}-\frac{1}{2t^2+1})dt=\arctan t-\int\frac{1}{2t^2+1}dt
\]
设 \(t=\frac{\sqrt{2}}{2}s,dt=\frac{\sqrt{2}}{2}ds\)
\[\int\frac{\sin^2x}{1+\sin^2x}dx=\arctan t-\frac{\sqrt{2}}{2}\int\frac{1}{s^2+1}ds=\arctan t-\arctan \sqrt{2}t=x-\arctan\sqrt{2}\tan x+C
\]
-
习题 4.2.2(7)
-
习题 4.2.2(8)
\[\int\frac{1}{\sin^4x\cos^4x}dx
\]
设 \(t=\tan x,dx=\frac{dt}{1+t^2}\)
\[\int\frac{1}{\sin^4x\cos^4x}dx=\int\frac{t^6+3t^4+3t^2+1}{t^4}dt=\int(t^2+3+\frac{3}{t^2}+\frac{1}{t^4})dt=\frac{t^3}{3}+3t-\frac{3}{t}-\frac{1}{3t^3}+C
\]
\[\int\frac{1}{\sin^4x\cos^4x}dx=\frac{\tan^3x}{3}+3\tan x-\frac{3}{\tan x}-\frac{1}{3\tan^3x}+C
\]
- 习题 4.2.2(9)
\[\int\frac{1}{2\sin x+\sin2x}dx=\frac{1}{2}\int\frac{1}{\sin x(1+\cos x)}dx
\]
设 \(t=\cos x,dx=-\frac{dt}{\sqrt{1=t^2}}\)
\[\int\frac{1}{2\sin x+\sin2x}dx=\int\frac{1}{\sqrt{1-t^2}(1+t)}\times(-\frac{1}{\sqrt{1-t^2}})dt=-\int\frac{1}{(1-t)(1+t)^2}=-\int(\frac{1}{2(1-t)}+\frac{t+1}{2(1+t)^2}-\frac{\frac{3}{2}}{(1+t)^2})dt=\frac{1}{2}\ln(1-t)-\frac{1}{4}\ln(1+t)^2+\frac{3}{2}\arctan t+C
\]
\[\int\frac{1}{2\sin x+\sin2x}dx=\frac{1}{2}\ln(1-\cos x)-\frac{1}{4}\ln(1+\cos x)^2+\frac{3}{2}\arctan\cos x+C
\]
- 习题 4.2.2(10)
\[\int\frac{\cos x}{a\sin x+b\cos x}dx
\]
设 \(t=\tan x,dx=\frac{dt}{1+t^2}\)
\[\int\frac{\cos x}{a\sin x+b\cos x}dx=\int\frac{1}{at+b}\times\frac{1}{1+t^2}dt
\]
待定系数解得
\[\int\frac{\cos x}{a\sin x+b\cos x}dx=\int(\frac{\frac{a}{a^2+b^2}}{at+b}+\frac{-\frac{t}{a^2+b^2}+\frac{1}{a(a^2+b^2)}}{t^2+1})dt=\frac{\ln(at+b)}{a^2+b^2}-\frac{\ln(t^2+1)}{2(a^2+b^2)}+\frac{\arctan t}{a(a^2+b^2)}+C
\]
\[\int\frac{\cos x}{a\sin x+b\cos x}dx=\frac{\ln(a\tan x+b)}{a^2+b^2}-\frac{\ln(a\tan x+b)}{2(a^2+b^2)}+\frac{x}{a(a^2+b^2)}+C
\]