Properties of Indefinite Integrals

Some properties of indefinite integrals

(i) The process of differentiation and integration are inverse of each other, i.e., \(\frac{d}{d x} \int f(x) d x=f(x)\) and \(\int f^{\prime}(x) d x=f(x)+\mathrm{C}\), where \(\mathrm{C}\) is any arbitrary constant.

Proof:

The process of differentiation and integration are inverses of each other in the sense of the following results:
Consider a function \(f\) such that its anti-derivative is given by \(F\), i.e.
\(
\frac{d}{d x} \mathrm{~F}(x)=f(x)
\)
Then,
\(
\int f(x) d x=\mathrm{F}(x)+\mathrm{C}
\)
On differentiating both the sides with respect to \(\mathrm{x}\) we have,
\(
\frac{d}{d x} \int f(x) d x=\frac{d}{d x}(\mathrm{~F}(x)+\mathrm{C})
\)
As we know, the derivative of any constant function is zero. Thus,
\(
\frac{d}{d x} \int f(x) d x=\frac{d}{d x}(\mathrm{~F}(x)+\mathrm{C})
\)
\(
\begin{aligned}
&=\frac{d}{d x} \mathrm{~F}(x) \\
&=f(x)
\end{aligned}
\)
The derivative of a function \(\mathrm{f}\) in \(\mathrm{x}\) is given as \(\mathrm{f}^{\prime}(\mathrm{x})\), so we get;
\(
f^{\prime}(x)=\frac{d}{d x} f(x)
\)
Therefore,
\(
\int f^{\prime}(x) d x=f(x)+\mathrm{C}
\)
Hence, proved.

(ii) Two indefinite integrals with the same derivative lead to the same family of curves and so they are equivalent. So if \(f\) and \(g\) are two functions such that \(\frac{d}{d x} \int f(x) d x=\frac{d}{d x} \int g(x) d x\), then \(\int f(x) d x\) and \(\int g(x) d x\) are equivalent.

Proof:

Let \(f\) and \(g\) be two functions such that
\(
\begin{gathered}
\frac{d}{d x} \int f(x) d x=\frac{d}{d x} \int g(x) d x \\
\text { or } \\
\frac{d}{d x}\left[\int f(x) d x-\int g(x) d x\right]=0
\end{gathered}
\)
Now,
\(
\begin{gathered}
\int f(x) d x-\int g(x) d x=\mathrm{C} \\
\int f(x) d x=\int g(x) d x+\mathrm{C}
\end{gathered}
\)
where \(\mathrm{C}\) is any real number.
Therefore, we cay say that, \(\int f(x) d x=\int g(x) d x\)

(iii) The integral of the sum of two functions equals the sum of the integrals of the functions i.e., \(\int(f(x)+g(x)) d x=\int f(x) d x+\int g(x) d x\).

Proof:

From the property (i) of integrals we have,
\(
\frac{d}{d x}\left[\int[f(x)+g(x)] d x\right]=f(x)+g(x) \ldots (1)
\)
Also, we can write;
\(
\frac{d}{d x}\left[\int f(x) d x+\int g(x) d x\right]=\frac{d}{d x} \int f(x) d x+\frac{d}{d x} \int g(x) d x=f(x)+g(x) \ldots (2)
\)
From (1) and (2),
\(
\int[f(x)+g(x)] d x=\int f(x) d x+\int g(x) d x
\)
Hence proved.

(iv) A constant factor may be written either before or after the integral sign, i.e., \(\int a f(x) d x=a \int f(x) d x\), where ‘ \(a\) ‘ is a constant.

Proof:

From property (i) we can say that
\(
\frac{d}{d x} \int a f(x) d x=a f(x)
\)
Also,
\(
\frac{d}{d x}\left[a \int f(x) d x\right]=a \frac{d}{d x} \int f(x) d x=a f(x)
\)
From property (ii) we can say that
\(
\int a f(x) d x=a \int f(x) d x
\)

(v) Properties (iii) and (iv) can be generalised to a finite number of functions \(f_1, f_2, \ldots, f_n\) and the real numbers, \(k_1, k_2, \ldots, k_n\) giving
\(
\int\left(k_1 f_1(x)+k_2 f_2(x)+\ldots+, k_n f_n(x)\right) d x=k_1 \int f_1(x) d x+k_2 \int f_2(x) d x+\ldots+k_n \int f_n(x) d x
\)

Example 1: Evaluate the given indefinite integral \(\int(6 x^5-18 x^2) d x.\)

Solution: Integrate the given function, it becomes:
\(
\int (6 x^5-18 x^2) d x=6\left(x^6 / 6\right)-18\left(x^3 / 3\right)+C
\)
Note: Don’t forget to put the integration constant ” \(\mathrm{C}\) “
After simplification, we get the solution
Thus, \(\int 6 x^5-18 x^2 d x=x^6-6 x^3+C\)

Example 2: Find the indefinite integral of \(\tan ^2 x\).

Solution:

We must find the integral of \(\tan ^2 x\).
\(
\begin{aligned}
&\int \tan ^2 x d x=\int\left(\sec ^2 x-1\right) d x \\
&=\int \sec ^2 x d x-\int 1 d x \\
&=\tan x-x+C
\end{aligned}
\)

Example 3:  Evaluate the following indefinite integral \(\int\left(\frac{2 a}{\sqrt{x}}-\frac{b}{x^2}+3 c \sqrt[3]{x^2}\right) d x\)

Solution:

\(
\begin{aligned}
&\int\left(\frac{2 a}{\sqrt{x}}-\frac{b}{x^2}+3 c \sqrt[3]{x^2}\right) d x \\
&=\int 2 a(x)^{\frac{-1}{2}} d x-\int b x^{-2} d x+\int 3 c x^{\frac{2}{3}} d x \\
&=4 a \sqrt{x}+\frac{b}{x}+\frac{9 c x^{\frac{5}{3}}}{5}+\mathrm{C}
\end{aligned}
\)

Example 4: Evaluate \(\int \frac{3 a x}{b^2+c^2 x^2} d x\)

Solution:

Let \(v=b^2+c^2 x^2\), then \(d v=2 c^2 x d x\)
Therefore, \(\int \frac{3 a x}{b^2+c^2 x^2} d x=\frac{3 a}{2 c^2} \int \frac{d v}{v}\)
\(
=\frac{3 a}{2 c^2} \log \left|b^2+c^2 x^2\right|+\mathrm{C} .
\)

Example 5: Verify the following using the concept of integration as an antiderivative.
\(\int \frac{x^3 d x}{x+1}=x-\frac{x^2}{2}+\frac{x^3}{3}-\log |x+1|+\mathrm{C}\)

Solution:

\(\frac{d}{d x}\left(x-\frac{x^2}{2}+\frac{x^3}{3}-\log |x+1|+\mathrm{C}\right)\)
\(
\begin{aligned}
&=1-\frac{2 x}{2}+\frac{3 x^2}{3}-\frac{1}{x+1} \\
&=1-x+x^2-\frac{1}{x+1}=\frac{x^3}{x+1} .
\end{aligned}
\)
Thus
\(\left(x-\frac{x^2}{2}+\frac{x^3}{3}-\log |x+1|+C\right)=\int \frac{x^3}{x+1} d x\)

Example 6: Evaluate \(\int \sqrt{\frac{1+x}{1-x}} d x, x \neq 1\)

Solution:

\(
\mathrm{I}=\int \sqrt{\frac{1+x}{1-x}} d x=\int \frac{1}{\sqrt{1-x^2}} d x+\int \frac{x d x}{\sqrt{1-x^2}}=\sin ^{-1} x+\mathrm{I}_1 \text {, }
\)
where \(\mathrm{I}_1=\frac{x d x}{\sqrt{1-x^2}}\).
Put \(1-x^2=t^2 \Rightarrow-2 x d x=2 t d t\). Therefore
\(
\mathrm{I}_1=-d t=-t+\mathrm{C}=-\sqrt{1-x^2}+\mathrm{C}
\)
Hence \(\quad I=\sin ^{-1} x-\sqrt{1-x^2}+C\).

Example 7: Evaluate \(\int \frac{d x}{\sqrt{(x-\alpha)(\beta-x)}}, \beta>\alpha\)

Solution: 

Put \(x-\alpha=t^2\). Then \(\beta-x=\beta-\left(t^2+\alpha\right)=\beta-t^2-\alpha=-t^2-\alpha+\beta\) and \(d x=2 t d t\). Now
\(
\begin{aligned}
&\mathrm{I}=\int \frac{2 t d t}{\sqrt{t^2\left(\beta-\alpha-t^2\right)}}=\int \frac{2 d t}{\sqrt{\left(\beta-\alpha-t^2\right)}} \\
&=2 \frac{d t}{\sqrt{k^2-t^2}}, \text { where } k^2=\beta-\alpha \\
&=2 \sin ^{-1} \frac{t}{k}+\mathrm{C}=2 \sin ^{-1} \sqrt{\frac{x-\alpha}{\beta-\alpha}}+\mathrm{C}
\end{aligned}
\)

Example 8: \(\text { Evaluate } \int \tan ^8 x \sec ^4 x d x\)

Solution: 

\(
\begin{aligned}
\mathrm{I} &=\int \tan ^8 x \sec ^4 x d x \\
&=\int \tan ^8 x\left(\sec ^2 x\right) \sec ^2 x d x \\
&=\int \tan ^8 x\left(\tan ^2 x+1\right) \sec ^2 x d x
\end{aligned}
\)
\(
\begin{aligned}
&=\int \tan ^{10} x \sec ^2 x d x+\int \tan ^8 x \sec ^2 x d x \\
&=\frac{\tan ^{11} x}{11}+\frac{\tan ^9 x}{9}+C
\end{aligned}
\)

Example 9: Evaluate \(\int \frac{x^3}{x^4+3 x^2+2} d x\)

Solution:

Put \(x^2=t\). Then \(2 x d x=d t\).
Now \(\quad \mathrm{I}=\int \frac{x^3 d x}{x^4+3 x^2+2}=\frac{1}{2} \int \frac{t d t}{t^2+3 t+2}\)
Consider \(\quad \frac{t}{t^2+3 t+2}=\frac{\mathrm{A}}{t+1}+\frac{\mathrm{B}}{t+2}\)
Comparing coefficient, we get \(\mathrm{A}=-1, \mathrm{~B}=2\).
Then \(\mathrm{I}=\frac{1}{2}\left[2 \int \frac{d t}{t+2}-\int \frac{d t}{t+1}\right]\)
\(
=\frac{1}{2}[2 \log |t+2|-\log |t+1|]
\)
\(
=\log \left|\frac{x^2+2}{\sqrt{x^2+1}}\right|+\mathrm{C}
\)

Example 10: \(\int \frac{d x}{2 \sin ^2 x+5 \cos ^2 x}\)

Solution: 

By dividing the numerator and denominator by \(\cos ^2 x\), we have
\(
\mathrm{I}=\int \frac{\sec ^2 x d x}{2 \tan ^2 x+5}
\)
Put \(\tan x=t\) so that \(\sec ^2 x d x=d t\). Then
\(
\begin{aligned}
\mathrm{I} &=\int \frac{d t}{2 t^2+5}=\frac{1}{2} \int \frac{d t}{t^2+\left(\sqrt{\frac{5}{2}}\right)^2} \\
&=\frac{1}{2} \frac{\sqrt{2}}{\sqrt{5}} \tan ^{-1}\left(\frac{\sqrt{2} t}{\sqrt{5}}\right)+\mathrm{C} \\
&=\frac{1}{\sqrt{10}} \tan ^{-1}\left(\frac{\sqrt{2} \tan x}{\sqrt{5}}\right)+\mathrm{C} .
\end{aligned}
\)