Integration problem 174,Ch.10 - Calculus by N. Piskunov

Question

I am trying to solve the problems in the section titled - Integration of irrational functions from the book Differential Calculus, N. Piskunov. My answer doesn't match that given in the text.

Could someone help me point, where I have made a mistake.

Evaluate the integral

$$I=\int{\sqrt{\frac{1-x}{1+x}}\frac{dx}{x^{2}}}$$

Solution. The given function is an irrational function $f(x) = \left(\frac{ax+b}{cx+d}\right)^\frac{1}{2}$.

We substitute,

$\displaystyle{\frac{1-x}{1+x}=t^{2}}$

We have:

$\displaystyle{x=\frac{1-t^{2}}{1+t^2}\\\frac{1}{x^2}=\left(\frac{1+t^2}{1-t^2}\right)^2}$

$\begin{align} dx&=\frac{(1+t^2)(-2t)-(1-t^2)(2t)}{(1+t^2)^2}dt\\ &=\frac{(-2t)(1+t^2+1-t^2)}{(1+t^2)^2}dt\\ &=\frac{(-4t)}{(1+t^2)^2}dt \end{align}$

We have:

$\begin{align} I&=\int{\sqrt{\frac{1-x}{1+x}}\frac{dx}{x^{2}}}\\ &=\int{t\cdot{\frac{-4t}{(1+t^2)^2}\cdot{\frac{(1+t^2)^2}{(1-t^2)^2}}}dt}\\ &=4\int{\frac{1-t^2-1}{(1-t^2)^2}dt}\\ &=4\int{\frac{dt}{(1-t^2)}}-4\int{\frac{dt}{(1-t^2)^2}}\\ &=4(I_{1}-I_{2}) \end{align}$

The first integral $I_{1}$:

$\begin{align} I_{1}&=\int{\frac{dt}{(1-t^2)}}\\ &=\frac{1}{2}\ln\left(\frac{1+t}{1-t}\right)+c_{1}\\ \end{align}$

The second integral $I_{2}$:

Let us write

$\begin{align} \frac{1}{(1-t)^2(1+t)^2}&=\frac{A}{1-t}+\frac{B}{(1-t)^2}+\frac{C}{1+t}+\frac{D}{(1+t)^2}\\ 1&=A(1-t)(1+t)^2+B(1+t)^2+C(1-t)^2(1+t)+D(1-t)^2 \end{align}$

When $t=1$, $4B=1$. $\implies{B=1/4}$.

When $t=-1$, $4D=1$. $\implies{D=1/4}$.

Also,

$1=(-A+C)t^3+(-A+B-C+D)t^2+(A+2B-C-2D)t+(A+B+C+D)$

$-A+C=0. \implies{A=C}$

$A+B+c+D=1. \implies{A=C=1/4}$

$\begin{align} I_{2}&=\frac{1}{4}\int{\frac{dt}{1-t}}+\frac{1}{4}\int{\frac{dt}{(1-t)^2}}+\frac{1}{4}\int{\frac{dt}{(1+t)}}+\frac{1}{4}\int{\frac{dt}{(1+t)^2}}\\ &=\frac{1}{4}\ln\left(\frac{1+t}{1-t}\right)+\frac{1}{4}\frac{1}{1-t}-\frac{1}{1+t}+c_{2}\\ &=\frac{1}{4}\ln\left(\frac{1+t}{1-t}\right)+\frac{1}{4}\frac{2t}{1-t^2}+c_{2} \end{align}$

Resulting integral $I$:

$\begin{align} I&=4(I_{1}-I_{2})\\ &=4\left[\frac{1}{4}\ln\left(\frac{1+t}{1-t}\right)-\frac{1}{4}\frac{2t}{1-t^2}+c\right]\\ &=\ln\left(\frac{\sqrt{1+x}+\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}\right)-\frac{2{\sqrt\frac{1+x}{1-x}}}{\frac{2x}{1+x}}+c \end{align}$

Answer 1

$$I=\int{\sqrt{\frac{1-x}{1+x}}\frac{dx}{x^{2}}}=\int{\sqrt{1-x^2}\over x^3+x^2}dx$$

$$\color {teal} {x=sin(u) => dx=cos(u)du}$$

$$\sqrt{1-x^2}=\sqrt{1-\sin^2u}=\cos u $$

$$\int {\cos^2u\over \sin^3 u+\sin^2u}du$$

$$\color {blue} {t=\tan({u\over 2}) => dt={1\over2}\sec^2({u\over2})du}$$ $$\sin u={2t\over t^2+1}\,\,\,\,\, ,\,\,\,\,\, \cos u={1-t^2\over t^2+1}\,\,\,\,\, ,\,\,\,\,\,du={2dt\over t^2+1}$$

$$\int {({1-t^2\over t^2+1})^2\over ({2t\over t^2+1})^3+({2t\over t^2+1})^2}{2dt\over t^2+1}=\int{(1-t^2)^2\over 4t^3+(t^2+1)2t^2}dt=\int{(-(t^2-1))^2\over 4t^3+2t^4+2t^2}dt=\int{(t^2-1)^2\over 2t^2(2t+t^2+1)}dt=\int{(t^2-1)^2\over 2t^2(t+1)^2}dt =\int{(t-1)^2(t+1)^2\over 2t^2(t+1)^2}dt$$$$=\int {(t-1)^2\over 2t^2}dt={1\over2}\int{(t-1)^2\over t^2}dt$$

$$\color {olive} {w=t-1 => dw=dt}$$

$${1\over2}\int{(t-1)^2\over t^2}dt={1\over2}\int({w\over w+1})^2dw={1\over2}\int({-2\over w+1 }+{1\over(w+1)^2}+1)dw$$$$={1\over2}(w - {1\over(w + 1)} - 2 \log(w + 1) + C)={1\over2}(t-{ 1\over t} - 2 \log(t) )+ C$$

$$\color {blue} {t=\tan({u\over 2})}$$

$${1\over2}(t-{ 1\over t} - 2 \log(t) )+ C={1\over2}(\tan({u\over 2})-{ 1\over \tan({u\over 2})} - 2 \log(\tan({u\over 2})) )+ C$$

$$\color{teal} {x=sin(u) => u=\arcsin x}$$

$${1\over2}(\tan({u\over 2})-{ 1\over \tan({u\over 2})} - 2 \log(\tan({u\over 2})) )+ C$$$$={1\over2}(\tan({\arcsin x\over 2})-{ 1\over \tan({\arcsin x\over 2})} - 2 \log(\tan({\arcsin x\over 2})) )+ C$$

$$\bbox[yellow,5px] {\tan({\arcsin x\over 2})= {x\over(\sqrt{1 - x} \sqrt{x + 1} + 1)}} \,\,\,\,\,\,\,\,\small{.(1)}$$

$${1\over2}(({x\over(\sqrt{1 - x} \sqrt{x + 1} + 1)} )-{ (\sqrt{1 - x} \sqrt{x + 1} + 1)\over x} - 2 \log({x\over(\sqrt{1 - x} \sqrt{x + 1} + 1)} ))+ C={1\over2}({-2 \sqrt{1 - x^2}\over x} - 2 \log({x\over(\sqrt{1 - x} \sqrt{x + 1} + 1)} ))+ C=({- \sqrt{1 - x^2}\over x} - \log({x\over(\sqrt{1 - x} \sqrt{x + 1} + 1)} ))+ C $$$$={- \sqrt{1 - x^2}\over x} +\log(\sqrt{1 - x^2}+ 1 )- \log{x} + C=({- \sqrt{1 - x^2}\over x} +\log{\sqrt{1 - x^2}+ 1 \over {x}} )+C $$

$$\bbox[yellow,5px,border:2px solid red]{=\ln\begin{vmatrix}{\sqrt{1-x}+\sqrt{1+x}\over\sqrt{1-x}-\sqrt {1+x}}\end{vmatrix}-{ \sqrt{1 - x^2}\over x}+C }$$

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⁽¹⁾

$\small{\tan {x\over 2}= {\sin x \over1+\cos x} }\,\,\,\,,\,\,\,\, \small{\cos x=\sqrt{1-\sin^2x}}$ $\small{\tan{\arcsin x\over 2}= {\sin \arcsin x \over1+\cos \arcsin x}={ x \over1+\cos \arcsin x}={ x \over1+\sqrt{1-[\sin\arcsin x]^2}}={ x \over1+\sqrt{1-x^2}}={ x \over1+\sqrt{1-x}\sqrt{1+x}} } $