Indefinite Integral Formulas

Integration of Indefinite Integral:

Integration is the reverse process of differentiation. In integration, we find the function whose differential coefficient is given. The rate of change of a quantity $y$ concerning another quantity $x$ is called the derivative or differential coefficient of $y$ concerning $x$. Geometrically, the Differentiation of a function at a point represents the slope of the tangent to the graph of the function at that point.

An intergral for which the limits are not defined is called the indefinite integral. An indefinite integral, denoted as $\int f(x) d x$. It is expressed as $\int f(x) d x=F(x)+C$. where $\mathrm{F}(\mathrm{x})$ is any antiderivative of $\mathrm{f}(\mathrm{x})$ and C is the constant of integration.

Fundamental Formulae (Inverse Trigonometric Functions)
1. $\frac{d}{d x}\left(\sin ^{-1} \frac{x}{a}\right)=\frac{1}{\sqrt{a^2-x^2}} \Rightarrow \int \frac{d x}{\sqrt{a^2-x^2}}=\sin ^{-1}\left(\frac{x}{a}\right)+C$
2. $\frac{d}{d x}\left(\cos ^{-1} \frac{x}{a}\right)=\frac{-1}{\sqrt{a^2-x^2}} \Rightarrow \int \frac{-1}{\sqrt{a^2-x^2}} d x=\cos ^{-1}\left(\frac{x}{a}\right)+C$
3. $\frac{d}{d x}\left(\frac{1}{a} \tan ^{-1} \frac{x}{a}\right)=\frac{1}{a^2+x^2} \Rightarrow \int \frac{d x}{a^2+x^2}=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+C$
4. $\frac{d}{d x}\left(\frac{1}{a} \cot ^{-1} \frac{x}{a}\right)=\frac{-1}{a^2+x^2} \Rightarrow \int \frac{-1}{a^2+x^2} d x=\frac{1}{a} \cot ^{-1}\left(\frac{x}{a}\right)+C$
5. $\frac{d}{d x}\left(\frac{1}{a} \sec ^{-1} \frac{x}{a}\right)=\frac{1}{x \sqrt{x^2-a^2}} \Rightarrow \int \frac{d x}{x \sqrt{x^2-a^2}}=\frac{1}{a} \sec ^{-1}\left(\frac{x}{a}\right)+C$
6. $\frac{d}{d x}\left(\frac{1}{a} \csc ^{-1} \frac{x}{a}\right)=\frac{-1}{x \sqrt{x^2-a^2}} \Rightarrow \int \frac{-d x}{x \sqrt{x^2-a^2}}=\frac{1}{a} \csc ^{-1}\left(\frac{x}{a}\right)+C$

Basic integration rules,
(i) $\frac{d}{d x}\left(\frac{x^{n+1}}{n+1}\right)=x^n$;
$\int x^n d x=\frac{x^{n+1}}{n+1}+$ C, $n \neq-1$

Particularly, we note that
$\frac{d}{d x}(x)=1$;
$\int d x=x+C$

(ii) $\frac{d}{d x}(\sin x)=\cos x$;
$\int \cos x d x=\sin x+C$

(iii) $\frac{d}{d x}(-\cos x)-\sin x$;
$\int \sin x d x=-\cos x+\mathrm{C}$

(iv) $\frac{d}{d x}(\tan x)-\sec ^2 x$;
$\int \sec ^2 x d x=\tan x+C$

(v) $\frac{d}{d x}(-\cot x)=\operatorname{cosec}^2 x$;
$\int \operatorname{cosec}^2 x d x=-\cot x+C$

(vi) $\frac{d}{d x}(\sec x)=\sec x \tan x$;
$\int \sec x \tan x d x=\sec x+C$

(vii) $\frac{d}{d x}(-\operatorname{cosec} x)=\operatorname{cosec} x \cot x$;
$\int \operatorname{cosec} x \cot x d x=-\operatorname{cosec} x+4$

(viii) $\frac{d}{d x}\left(\sin ^{-1} x\right)-\frac{1}{\sqrt{1-x^2}}$;
$\int \frac{d x}{\sqrt{1-x^2}}=\sin ^{-1} x+C$

(ix) $\frac{d}{d x}\left(-\cos ^{-1} x\right)=\frac{1}{\sqrt{1-x^2}}$;
$\int \frac{d x}{\sqrt{1-x^2}}=-\cos ^{-1} x+C$

(x) $\frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^2}$;
$\int \frac{d x}{1+x^2}=\tan ^{-1} x+C$

(xi) $\frac{d}{d x}\left(-\cot ^{-1} x\right)=\frac{1}{1+x^2} ; \quad \int \frac{d x}{1+x^2}=-\cot ^{-1} x+\mathrm{C}$

Recommended Videos Based on Indefinite Integral

Solved Examples based on Indefinite Integral

Example 1: If $\int \frac{2 x+5}{\sqrt{7-6 x-x^2}} d x=A \sqrt{7-6 x-x^2}+B \sin ^{-1}\left(\frac{x+3}{4}\right)+C$ (where C is a constant of integration), then the ordered pair ( A , B ) is equal to :
1) $(2,1)$
2) $(-2,-1)$
3) $(-2,1)$
4) $(2,-1)$

Solution
Integration by perfect square method -

$a x^2+b x+c=a\left[x^2+\frac{b x}{a}+\frac{c}{a}\right]=a\left[\left(x+\frac{b}{2 a}\right)^2+\frac{c}{a}-\frac{b^2}{4 a^2}\right]$

Make the coefficient of $x^2$ +ve one .

and

$\begin{aligned}
& I=\int \frac{2 x+5}{\sqrt{7-6 x-x^2}} d x \\
& =-\int \frac{(-2 x-5)}{\sqrt{7-6 x-x^2}} d x \\
& =-\int \frac{-6-2 x-5+6}{\sqrt{7-6 x-x^2}} d x \\
& =\int\left(-\frac{-2 x-6}{\sqrt{-x^2-6 x+7}}-\frac{1}{\sqrt{-x^2-6 x+7}}\right) \mathrm{d} x \\
& =-2 \sqrt{-x^2-6 x+7}-\int \frac{1}{\sqrt{16-(x+3)^2}} \mathrm{~d} x \\
& =-2 \sqrt{7-6 x-x^2}-\sin ^{-1}\left(\frac{x+3}{4}\right)+C \\
& \mathrm{~A}=-2 ; \mathrm{B}=-1
\end{aligned}$

Hence, the answer is the option 2.

Example 2: If $I_1=\int \frac{d x}{\sqrt{a^2-x^2}}$ and $I_2=\int \cos ^{-1} \frac{x}{a}$ then $I_1+I_2=$ ?

1) $\sin ^{-1} \frac{x}{a}+c$

2) $\cos ^{-1} \frac{x}{a}+c$

3) Constant value

4) None of these

Solution

As we have learnt,

Integrals of inverse circular functions -

$\begin{aligned}
& \frac{\mathrm{d}}{\mathrm{d} x}\left(\sin ^{-1} \frac{x}{a}\right)=\frac{1}{\sqrt{a^2-x^2}} \\
& \therefore \int \frac{d x}{\sqrt{a^2-x^2}}=\sin ^{-1} \frac{x}{a}+c \\
& I_1=\int \frac{d x}{\sqrt{a^2-x^2}}=\sin ^{-1} \frac{x}{a}+c
\end{aligned}$

Also,

$I_1+I_2=\sin ^{-1} \frac{x}{a}+\cos ^{-1} \frac{x}{a}+c=\frac{\pi}{2}+c=C$

Hence, the answer is the option 3.

Example 3: If the value of the integral $\int_0^{\frac{1}{2}} \xrightarrow[\left(1-x^2\right)^{\frac{3}{2}}]{x^2} d x$ is $\frac{k}{6}$ then $k$ is equal to?
1) $2 \sqrt{3}-\pi$
2) $2 \sqrt{3}+\pi$
3) $3 \sqrt{2}+\pi$
4) $3 \sqrt{2}-\pi$

Solution

$\begin{aligned}
& \int_0^{1 / 2} \frac{\left(\left(x^2-1\right)+1\right)}{\left(1-x^2\right)^{3 / 2}} d x \\
& \int_0^{1 / 2} \frac{d x}{\left(1-x^2\right)^{3 / 2}}-\int_0^{1 / 2} \frac{d x}{\sqrt{1-x^2}} \\
& \int_0^{1 / 2} \frac{x^{-3}}{\left(x^{-2}-1\right)^{3 / 2}} d x-\left(\sin ^{-1} x\right)_0^{1 / 2}
\end{aligned}$

Let $x^{-2}-1=t^2 \Rightarrow x^{-3} d x=-t d t$

$\begin{aligned}
& \int_{\infty}^{\sqrt{3}} \frac{-t d t}{t^3}-\frac{\pi}{6}=\int_{\sqrt{3}}^{\infty} \frac{d t}{t^2}-\frac{\pi}{6}=\frac{1}{\sqrt{3}}-\frac{\pi}{6}=\frac{k}{6} \\
& k=2 \sqrt{3}-\pi
\end{aligned}$
Hence, the answer is the option 1 .

Example 4: Let $\mathrm{y}=\mathrm{y}(\mathrm{x})$ be the solution of the differential equation $x d y-y d x=\sqrt{\left(x^2+y^2\right)} d x, x \geqslant 1$, with $\mathrm{y}(1)=0$. If the area bounded by the line $\mathrm{x}=1, x=e^\pi, y=0$ and $\mathrm{y}=\mathrm{y}(\mathrm{x})$ is $\alpha e^{2 \pi}+\beta$, then the va;ue of $10(\alpha+\beta)$ is equal to $\qquad$
1) $2$
2) $4$
3) $6$
4) $8$

Solution

$\begin{aligned}
& x d y-y d x=\sqrt{x^2-y^2} d x \\
& \Rightarrow \int \frac{x d y-y d x}{x^2}=\frac{1}{x} \sqrt{1-\frac{y^2}{x^2}} d x \\
& \Rightarrow \int \frac{d\left(\frac{y}{x}\right)}{\sqrt{1-\left(\frac{y}{x}\right)^2}}=\int \frac{d x}{x} \\
& \Rightarrow \sin ^{-1}\left(\frac{y}{x}\right)=\ln |x|+c \\
& \text { at } x=1, y=0 \Rightarrow c=0 \\
& y=x \sin (\ln x)
\end{aligned}$

$\begin{aligned}& \mathrm{A}=\int_1^{\mathrm{e}^\pi} \mathrm{x} \sin (\ell \mathrm{nx}) \mathrm{dx} \\
& \mathrm{x}=\mathrm{e}^{\mathrm{t}}, \mathrm{dx}=\mathrm{e}^{\mathrm{t}} d \mathrm{dt} \\
& \Rightarrow \int_0^{\pi^2} \mathrm{e}^{2 \mathrm{t}} \sin (\mathrm{t}) \mathrm{dt}=\mathrm{A} \\
& \alpha \mathrm{e}^{2 \pi}+\beta=\left(\frac{\mathrm{e}^{2 t}}{5}(2 \sin \mathrm{t}-\cos \mathrm{t})\right)_0^\pi=\frac{1+\mathrm{e}^{2 \pi}}{5} \\
& \alpha=\frac{1}{5}, \beta=\frac{1}{5} \text { so } 10(\alpha+\beta)=4
\end{aligned}$
Hence, the answer is the option 2.

Example 5: Evaluate the integral of $\int \frac{1}{5 \sqrt{25-4 x^2}} d x$.
1) $\frac{1}{10} \sin ^{-1} \frac{2 x}{5}+C$
2) $-\frac{1}{4 x} \sec ^{-1} \frac{5}{2 x}+C$
3) $\frac{1}{5} \sec ^{-1} \frac{4 x}{5}+C$
4) $-\frac{1}{5} \sec ^{-1} \frac{4 x}{5}+C$

Solution

Given integral

$\int \frac{1}{5 \sqrt{25-4 x^2}} d x$

Substitute $x=\frac{5}{2} \sin \theta$ and $d x=\frac{5}{2} \cos \theta d \theta$.
$\theta=\sin ^{-1} \frac{2 x}{5}$
$\int \frac{1}{5 \sqrt{25-4 x^2}} d x=\int \frac{1}{5 \sqrt{25-\left(4 \times\left(\frac{5}{2} \sin \theta\right)^2\right)}} \frac{5}{2} \cos \theta d \theta$
$\int \frac{1}{\sqrt{25\left(1-\sin ^2 \theta\right)}} \frac{1}{2} \cos \theta d \theta=\frac{1}{2} \int \frac{1}{5 \cos \theta} \times \cos \theta d \theta$
$\frac{1}{10} \int d \theta=\frac{1}{10} \theta+C$
So, it becomes
$\frac{1}{10} \sin ^{-1} \frac{2 x}{5}+C$

Hence, the answer is option (1).

Summary

Indefinite integration is the most elementary process in calculus for seeking antiderivatives of functions. It reverses the process of differentiation, which finds a function whose derivative equals the original function plus an arbitrary constant. Basic rules of integration allow integration with polynomial functions, exponential functions, trigonometric functions, and so on.