Integration using variable substitution

Integration by substitution (change of variable).

Let a composite function $f(\varphi (x))$ be defined on some interval, and let the function $t=\varphi (x)$ 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 $\int f(\varphi (x)){\varphi }^{\prime }(x),dx$ also exists, and moreover

$$\int f(\varphi (x)){\varphi }^{\prime }(x)dx={\int f(t)dt|}_{t=\varphi (x)}.(1)$$

This formula is called the substitution integration formula.

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

$$\int f(t)dt={\int f(\varphi (x)){\varphi }^{\prime }(x)dx|}_{x={\varphi }^{-1}(t)},$$ or, if we denote the original integration variable as usual by $x,$ $$\int f(x)dx={\int f(\varphi (t)){\varphi }^{\prime }(t)dt|}_{t={\varphi }^{-1}(x)}.(2)$$

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

Examples:

Compute the integrals using an appropriate substitution:

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

Solution.

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

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

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

Solution.

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

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

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

Solution.

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

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

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

Solution.

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

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

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

Solution.

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

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

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

Solution.

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

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

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

Solution.

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

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

Applying the specified substitutions, find the integrals:

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

Solution.

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

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

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

Solution.

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

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

Using suitable substitutions, find the integrals:

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

Solution.

$$\int x(5x-1{)}^{1}9dx=\left[\begin{array}{l}x=\frac{t+1}{5}\\ dx=\frac{1}{5}dt\end{array}\right]=\int \frac{t+1}{5}{t}^{19}\frac{dt}{5}=$$ $$=\frac{1}{25}\int (t^{20}+t^{19})dt=$$ $$=\frac{1}{25}(\frac{t^{21}}{21}+\frac{t^{20}}{20})+c=\frac{1}{25}(\frac{(5x-1{)}^{21}}{21}+\frac{(5x-1{)}^{20}}{20})+c.$$

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

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

Solution.

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

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

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

Solution.

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

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

Homework.

Compute the integrals using an appropriate substitution:

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

2. $\int chxshxdx.$

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

4. $\int ctgxdx.$

5. $\int \cos (ax+b)dx.$

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

7. $\int \frac{dx}{\cos (x-\frac{\pi }{4})}.$

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

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

10. $\int \frac{\sin xdx}{\sqrt{{\cos }^{2}x+4}}.$

11. $\int \frac{dx}{4x^2+7}.$

12. $\int \frac{xdx}{4x^2+7}.$

Applying the specified substitutions, find the integrals:

13. $\int \frac{dx}{x\sqrt{4-x^2}},x=\frac{2}{t}.$

14. $\int \frac{e^{2x}}{e^x+1},x=\ln t.$

Using suitable substitutions, find the integrals:

15. $\int \frac{e^{3x}}{\sqrt{1-e^x}}dx.$

16. $\int \frac{xdx}{(3-x{)}^7}.$

17. $\int \frac{dx}{x\sqrt{x^2+1}}.$

Tags: calculus, integral, mathematical analysis, variable substitution