This section consider the Method of Substitution to evaluate the integration. The process is given in details. The problems listed here can not be solved directly using Antiderivatives; but need to apply Method of Substitution before using Antiderivatives.
Rule of Method of Substitution
If \( u = g(x) \) is a differentiable function whose range is an interval I , and f is continuous on I , then \[ \int g'(x) f(g(x)) dx = \int f(u) du. \]The rule apply to solve examples as below
Example by Method of Substitution
Example involving Algebraic Expressions
To apply the substitution method, you need to identify the function \(g(x)\) by observing the given problem, and during selection of \(g(x)\) , keep in mind that after substitution the integration should reduce to the form of Antiderivative table format.
Q. Evaluate \(\int x \sqrt{x^2 + 5 } dx \).
Ans. Follow the below steps
- Thought Process for the selection of \(g(x)\) : From the table of Antiderivative we know the integration of \( \int \sqrt{x} dx \), but there is none about \( \int \sqrt{x^2 + a} dx \). Thus, chose \(\underline{u = x^2 +5}\) .
- Take the derivative w.r.t \(x\) on both side : \[ du = 2 x dx \implies x dx = \frac{du}{2} . \] Remark: Your target here to find a relation between \(du\) and \(dx\). For writing of solution you can write directly as written above but for knowledge you must aware the background process which is influenced by the Chain Rule as follows: \[ du = \frac{du}{dx} dx \] In this problem this is given as \[ du = \frac{du}{dx} dx = \frac{d (x^2+5) }{dx} dx = 2 x dx \]
- Replace \(x\) by new variables : The given integration changes to (Antiderivative table form) \[\int x \sqrt{x^2 + 5 } dx = \int \sqrt{\color{red}{x^2 + 5} }\; {\color{blue}{x dx}} =\int \sqrt{u}\; \frac{du}{2} =\frac{1}{2} \bigg(\frac{u^{3/2}}{3/2} \bigg) + C = \frac{1}{3} u^{3/2} + C. \]
- Final Step: Replacement to original variable : To have the final answer $u$ should be replace back by $x^2+5$. Thus, finally \[\int x \sqrt{x^2 + 5 } dx = \frac{1}{3} \left(x^2+5\right)^{3/2} + C. \]
Q. Evaluate \(\int (4 x^2+ 3x)^{25} (16x+6) dx \).
Ans. Follow the below steps
- Take $u = 4 x^2+ 3x$
- Then $du = (8x +3) dx $
- This implies $du = (8x +3) dx. $
- Thus, \[\int (4 x^2+ 3x)^{25} (16x+6) dx = 2 \int (4 x^2+ 3x)^{25} (8 x+3) dx = \int (u)^{25} du = 2 \frac{u^{26}}{26} + c \]
- Finally, \[\int (4 x^2+ 3x)^{25} (16x+6) dx = \frac{(4 x^2+ 3x)^{26}}{13} + c \]
Q. Evaluate \(\int x^2 \sqrt{x + 5 } dx \).
Ans. Follow the below steps
- Put $u=x+5 \implies du = dx $.
- This gives $x=u-5$ and the given integration changes to \[\begin{align*} \int (u-5)^2 \sqrt{u} \sqrt{u } du &= \int (u^2-10 u+ 25) \sqrt{u} du \\ &= \int (u^{\frac{5}{2}}-10 u^{\frac{3}{2}}+ 25 u^{\frac{1}{2}}) du \\ &= \frac{u^{\frac{7}{2}}}{\frac{7}{2}}-10 \frac{u^{\frac{3}{2}}}{\frac{3}{2}}+ 25 \frac{u^{\frac{1}{2}}}{\frac{1}{2}} +c \\ &= \frac{2}{7} (x+5)^{\frac{7}{2}}-\frac{20}{3} (x+5)^{\frac{3}{2}} + 50 (x+5)^{\frac{1}{2}} + c \end{align*}\]
Example involving trigonometric functions
Q. Evaluate \(\int \sec^2(2x)\sqrt{\tan(2x) } dx \).
Ans. Follow the below steps
- Substitute \(u = \tan(2x) \).
- Then, \(du = 2 \sec^2(2x) dx \implies \sec^2(2x) dx = \frac{du}{2}\).
- Thus, \[\int \sec^2(2x)\sqrt{\tan(2x) } dx = \frac{1}{2} \int \sqrt{u} du = \frac{1}{2} \frac{u^{3/2}}{3/2} + c= \frac{1}{3} u^{3/2} + c. \]
- Finally, we have \[ \int \sec^2(2x)\sqrt{\tan(2x) } dx = \frac{1}{3} \tan^{3/2}(2x) + c . \]
Q. Evaluate \(\int \tan(x) dx \).
Ans. There is no function whose derivative is $\tan(x)$. So, rewrite the integration as \(\int \frac{\sin(x)}{\cos(x)} dx. \) Then follow the below steps
- Substitute \(u = \cos(x) \).
- Then, \(du = – \sin(x) dx \implies \sin() dx = – du\).
- Thus, \[\int \frac{\sin(x)}{\cos(x)} dx = -\int \frac{1}{u} du = – \ln(u) + c. \]
- Finally, we have
\[ \int \tan(x) dx = – \ln(\cos(x))+ c = \ln(\sec(x))+c. \]
\(- \ln(\cos(x))= \ln(cos(x))^{-1}= \ln\bigg(\frac{1}{\cos(x)}\bigg)=\ln(sec(x)) \)
Note: Similarly \(\int \cot(x) dx = \ln( \sin(x) ) +c \)
Example involving exponential and logarithm
Q. Evaluate \(\int \frac{e^x}{e^{2x}+4} dx \).
Ans. Follow the below steps
- Substitute $t = e^{x}$,
- This implies $dt = e^x dx$.
- By this the integration \(\int \frac{e^x}{e^{2x}+4} dx \) reduces to \[\int \frac{1}{t^2+4} dt = \frac{1}{2} \tan^{-1}\left(\frac{t}{2}\right)+c\]
- Finally, we have \[\int \frac{e^x}{e^{2x}+4} dx = \frac{1}{2} \tan^{-1}\left(\frac{e^{x}}{2}\right) + c . \]
Note: Here we are using \(\int \frac{1}{x^2+a^2}dx = \frac{1}{a} \tan^{-1}\bigg(\frac{x}{a}\bigg)+c. \)
Q. Evaluate \(\int \frac{1}{e^{3x}+ e^{-3x}} dx \).
Ans. Follow the below steps
- Substitute $t = e^{3x}$,
- This implies $dt = 3 e^{3x} dx \implies e^{3x} dx = \frac{dt}{3}$.
- First, we need to rearrange \(\int \frac{1}{e^{3x}+ e^{-3x}} dx \) as \[\int \frac{1}{e^{3x}+ e^{-3x}} dx = \int \frac{e^{3x}}{e^{6x}+ 1} dx \] Then, it reduces to \[\int \frac{1}{2}\frac{1}{t^2+1}dt = \frac{1}{2} \tan^{-1}\left(t\right)+c\]
- Finally, we have \[\int \frac{1}{e^{3x}+ e^{-3x}} dx = \frac{1}{2} \tan^{-1}\left(e^{3x}\right) +c . \]
> Q. Evaluate \(\int \frac{1}{x \ln(x)} dx \).
Ans. Follow the below steps
- Substitution : Since, we do not know any function, whose derivative gives us $\ln$ or any function involving $\ln$ . Hence, we have to substitute such that $\ln$ vanish.
In this problem, let $t= \ln(x)$ . Then \[ dt = \frac{1}{x} dx. \] - The given integration reduces to \[ \int \frac{1}{t} dt = \ln(t) + c = \ln(\ln(x))+c. \]