Integration using variable substitution

Literature: Collection of problems in mathematics. Part 1. Edited by A. V. Efimov, B. P. Demidovich.

Integration by substitution (change of variable).

Let a composite function $f(\phi(x))$ be defined on some interval, and let the function $t=\phi(x)$ be continuous on this interval and differentiable at all of its interior points; then if the integral $\int f(t), dt$ exists, then the integral $\int f(\phi(x))\phi'(x),dx$ also exists, and moreover

$$\int f(\phi(x))\phi'(x)\,dx=\left.\int f(t)\,dt\right|_{t=\phi(x)}.\qquad\qquad (1)$$

This formula is called the substitution integration formula.

If there exists an inverse function $x=\phi^{-1}(t)$ for the function $t=\phi(x)$ on the interval under consideration, then formula (1) can be rewritten as

$$\int f(t)\,dt=\left.\int f(\phi(x))\phi'(x)\,dx\right|_{x=\phi^{-1}(t)},$$ or, if we denote the original integration variable as usual by $x,$ $$\int f(x)\,dx=\left.\int f(\phi(t))\phi'(t)\,dt\right|_{t=\phi^{-1}(x)}.\qquad\qquad (2)$$

Formula (2) is usually called the variable substitution integration formula.

Examples:

Compute the integrals using an appropriate substitution:

6.44. $\int\sqrt{3+x},dx.$

Solution.

$$\int\sqrt{3+x}\,dx=\left[\begin{array}{lcl}t=3+x\\ dt=dx\end{array}\right]=\int\sqrt{t}dt=\int t^{\frac{1}{2}}dt=\frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}+c=$$ $$=\frac{2t^{3/2}}{3}+c=\frac{2\sqrt{(3+x)^3}}{3}+c.$$

Answer: $\frac{2\sqrt{(3+x)^3}}{3}+c.$

6.48.$\int\frac{dx}{x\ln^2 x}.$

Solution.

$$\int\frac{dx}{x\ln^2 x}=\left[\begin{array}{lcl}t=ln x\\ dt=\frac{dx}{x}\end{array}\right]=\int\frac{dt}{t^2}=\int t^{-2}dt=\frac{t^{-2+1}}{-2+1}+c=$$ $$=\frac{t^{-1}}{-1}=-\frac{1}{t}+c=-\frac{1}{\ln x}+c.$$

Answer: $-\frac{1}{\ln x}+c.$

6.49.$\int\frac{dx}{a+bx}.$

Solution.

$$\int\frac{dx}{a+bx}=\left[\begin{array}{lcl}t=a+bx\\ dt=b{dx}\end{array}\right]=\int\frac{dt}{bt}=\frac{1}{b}\int \frac{dt}{t}=\frac{1}{b}\ln|t|+c=$$ $$=\frac{1}{b}\ln|a+bx|+c.$$

Answer: $\frac{1}{b}\ln |a+bx|+c.$

6.55.$\int\sin(\ln x)\frac{dx}{x}.$

Solution.

$$\int\sin(\ln x)\frac{dx}{x}=\left[\begin{array}{lcl}t=ln x\\ dt=\frac{dx}{x}\end{array}\right]=\int\sin t\,dt=-\cos t+c=\cos\ln x+c.$$

Answer: $\cos\ln x+c.$

6.59.$\int\frac{xdx}{\sqrt[3]{x^2-1}}.$

Solution.

$$\int\frac{xdx}{\sqrt[3]{x^2-1}}=\left[\begin{array}{lcl}t=x^2-1\\ dt=2xdx\Rightarrow xdx=\frac{dt}{2} \end{array}\right]=\int\frac{dt}{2\sqrt[3]{t}}=$$ $$=\frac{1}{2}\int t^{-1/3}dt+c=\frac{1}{2}\frac{t^{-1/3+1}}{-1/3+1}+c=\frac{1}{2}\frac{t^{2/3}}{2/3}+c=\frac{3}{4}\sqrt[3]{t^2}+c=$$ $$=\frac{3\sqrt[3]{(x^2-1)^2}}{4}+c.$$

Answer: $\frac{3\sqrt[3]{(x^2-1)^2}}{4}+c.$

6.69.$\int\frac{\sin ax}{\cos^3 ax}\,dx.$

Solution.

$$\int\frac{\sin ax dx}{\cos^3 ax}=\left[\begin{array}{lcl}t=\cos ax\\ dt=-a\sin ax dx\Rightarrow \sin ax d=-\frac{dt}{a}\end{array}\right]=-\int\frac{dt}{at^3}=$$ $$=-\frac{1}{a}\int t^{-3}dt=-\frac{1}{a}\frac{t^{-3+1}}{-3+1}+c=-\frac{t^{-2}}{-2a}+c=\frac{1}{2at^2}+c=\frac{1}{2a\cos^2 ax}+c.$$

Answer: $\frac{1}{2a\cos^2 ax}+c.$

6.71.$\int\frac{e^x}{(7-e^x)^2}\,dx.$

Solution.

$$\int\frac{e^x\,dx}{7-e^x}=\left[\begin{array}{lcl}t=7-e^x\\ dt=-e^x\,dx\end{array}\right]=-\int\frac{dt}{t}=-\ln |t|+c=-\ln|7-e^x|+c.$$

Answer: $-\ln|7-e^x|+c.$

Applying the specified substitutions, find the integrals:

6.114. $\int\frac{dx}{x\sqrt{1-x^3}},\quad x=(1-t^2)^{1/3}.$

Solution.

$$\int\frac{dx}{x\sqrt{1-x^3}}=\left[\begin{array}{lcl}x=(1-t^2)^{1/3}\\ dx=\frac{1}{3}(1-t^2)^{-2/3}(-2t)dt\end{array}\right]=$$ $$=\int\frac{\frac{-2t}{3(1-t^2)^{2/3}}dt}{(1-t^2)^{1/3}\sqrt{1-(1-t^2)}}=\int\frac{-2tdt}{3(1-t^2)^{2/3}(1-t^2)^{1/3}t}=$$ $$=-2\int\frac{dt}{3(1-t^2)}=-\frac{2}{3}\left(\frac{1}{2}\int\frac{dt}{1-t}+\frac{1}{2}\int\frac{dt}{1+t}\right)=$$ $$=-\frac{1}{3}(-\ln|1-t|+\ln|1+t|)+c=\frac{1}{3}\ln\left|\frac{1-\sqrt{1-x^3}}{1+\sqrt{1-x^3}}\right|+c.$$

Answer: $\frac{1}{3}\ln\left|\frac{1-\sqrt{1-x^3}}{1+\sqrt{1-x^3}}\right|+c.$

6.116.$\int\frac{dx}{x+\sqrt {x}},\quad x=t^2.$

Solution.

$$\int\frac{dx}{x+\sqrt{x}}=\left[\begin{array}{lcl}x=t^2\\ dx=2tdt\end{array}\right]=\int\frac{2tdt}{t^2+t}=2\int\frac{dt}{t+1}=$$ $$=2\ln|t+1|+c=2\ln(\sqrt {x}+1)+c.$$

Solution: $2\ln(\sqrt{x}+1)+c.$

Using suitable substitutions, find the integrals:

6.118.$\int x(5x-1)^{19}\, dx.$

Solution.

$$\int x(5x-1)^19\,dx=\left[\begin{array}{lcl}x=\frac{t+1}{5}\\ dx=\frac{1}{5}dt\end{array}\right]=\int\frac{t+1}{5}t^{19}\frac{dt}{5}=$$ $$=\frac{1}{25}\int(t^{20}+t^{19})\,dt=$$ $$=\frac{1}{25}\left(\frac{t^{21}}{21}+\frac{t^{20}}{20}\right)+c=\frac{1}{25}\left(\frac{(5x-1)^{21}}{21}+\frac{(5x-1)^{20}}{20}\right)+c.$$

Answer: $\frac{1}{25}\left(\frac{(5x-1)^{21}}{21}+\frac{(5x-1)^{20}}{20}\right)+c.$

6.120.$\int\frac{x+2}{\sqrt{x+1}+1}\,dx.$

Solution.

$$\int \frac{x+2}{\sqrt{x+1}+1}\,dx=\left[\begin{array}{lcl}x=t^2-1\\ dx=2tdt\end{array}\right]=\int\frac{t^2+1}{t+1}2t\,dt=$$ $$=2\int\frac{t^3+1}{t+1}\,dt=2\int(t^2-t+1)\,dt=2\left(\frac{t^3}{3}-\frac{t^2}{2}+t\right)+c=$$ $$=2\left(\frac{\sqrt{(x+1)^3}}{3}+\frac{\sqrt{(x+1)^2}}{2}+\sqrt{x+1}\right)+c=$$ $$=2\left(\frac{\sqrt{(x+1)^3}}{3}+\frac{x+1}{2}+\sqrt{x+1}\right)+c.$$

Answer: $2\left(\frac{\sqrt{(x+1)^3}}{3}+\frac{x+1}{2}+\sqrt{x+1}\right)+c.$

6.122.$\int\frac{dx}{\sqrt{3+e^x}}.$

Solution.

$$\int \frac{dx}{\sqrt{3+e^x}}\,dx=\left[\begin{array}{lcl}x=\ln(t^2-3)\\ dx=\frac{2t}{t^2-3}\end{array}\right]=\int\frac{2t}{(t^2-3)t}\,dt=2\int\frac{1}{t^2-3}\,dt=$$ $$=2\frac{1}{2\sqrt{3}}\ln\left|\frac{x-\sqrt {3}}{x+\sqrt{3}}\right|+c=\frac{1}{\sqrt {3}}\ln\left|\frac{x-\sqrt{3}}{x+\sqrt{3}}\right|.$$

Answer: $\frac{1}{\sqrt{3}}\ln\left|\frac{x-\sqrt{3}}{x+\sqrt{3}}\right|+c.$

Homework.

Compute the integrals using an appropriate substitution:

6.45. $\int(3-4\sin x)^{1/3}\cos x\,dx.$

6.46. $\int ch x sh x\, dx.$

6.51. $\int\frac{\cos\frac{x}{\sqrt 2}}{2-3\sin\frac{x}{\sqrt 2}}\, dx.$

6.52. $\int ctg x\, dx.$

6.54. $\int\cos(ax+b)\, dx.$

6.56. $\int\sin\sqrt{x}\frac{dx}{\sqrt x}.$

6.57. $\int\frac{dx}{\cos\left(x-\frac{\pi}{4}\right)}.$

6.62. $\int\frac{e^{-\alpha x}}{1+e^{-2\alpha x}}\, dx.$

6.63. $\int\frac{dx}{\sqrt{5-3x^2}}.$