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Algebraic Integrals

Finding the solution to algebraic integrals is a lengthy and complex process. Hence, a standard method is to be followed to attain the solution in an easy way. In this section, we will discuss different forms of algebraic integrals, and tricks given here will help aspirants to solve all the problems at their own pace.

The types of integrals are as follows:

  • The integral can be called a definite integral if the function f(x) is between the limits β€˜a’ and β€˜b’, where d/dx (F(x)) = f(x).
It is represented as follows:

\(\begin{array}{l}\int_{a}^{b}f(x)\ dx = F(b) – F(a)\end{array} \)
  • Indefinite integrals refer to the evaluation of the indefinite area. The diagram below explains the difference between definite and indefinite integral.

Difference between definite and indefinite integrals

Algebraic Integrals Forms

The following forms of integrals can be categorised into algebraic integrals.

1]

\(\begin{array}{l}\int \frac{1}{ax^2+bx+c}dx\ \text{or reducible to}\ \int \frac{1}{ax^2+bx+c}dx\end{array} \)

Tricks to Solve: In these integrals, try expressing the term ax2 + bx + c as the sum or difference of the two squares. The following steps can be followed:

  • By dividing or multiplying throughout, the coefficient of x2 can be made unity.
  • \(\begin{array}{l}\text{To express it in the form}\ [x+\frac{b}{2a}]^2 + \frac{4ac-b^2}{4a^2},\end{array} \)
    add and subtract the square of half of the coefficient of x.
  • This reduces to one of the forms and can be integrated easily.
    \(\begin{array}{l}\frac{1}{x^2-a^2}\ \text{or}\ \frac{1}{x^2+a^2}\ \text{or}\ \frac{1}{a^2-x^2}\end{array} \)

2]

\(\begin{array}{l}\int \frac{1}{\sqrt{ax^2+bx+c}}dx\ \text{or reducible to}\ \int \frac{1}{\sqrt{ax^2+bx+c}}dx\end{array} \)

Tricks to Solve: To integrateΒ these functions, the following steps can be followed:

  • Make the coefficient of x2 as unity.
  • To express the term within the square root in the form
    \(\begin{array}{l}[x+\frac{b}{2a}]^2 + \frac{4ac-b^2}{4a^2},\end{array} \)
    , add and subtract the square of half of the coefficient of x.
  • This reduces to one of the forms and can be integrated easily.
    \(\begin{array}{l}\frac{1}{x^2-a^2}\ or\ \frac{1}{x^2+a^2}\ or\ \frac{1}{a^2-x^2}\end{array} \)

3]

\(\begin{array}{l}\int \frac{px+q}{ax^2+bx+c}\end{array} \)

Tricks to Solve:

  • In such cases, the numerator can be expressed as the sum of constant times differentiation of the denominator and a constant; that is, px + q is written as A (2ax + b) + C, where 2ax + b= d/dx (ax2+bx+c). A and C are arbitrary constants.
  • By equating the coefficients of like powers on both sides, the values of the constants can be obtained easily.
  • Replace the numerator (px + q) by A (2ax + b) + C in the given integral, which will give,
    \(\begin{array}{l}\int\frac{px+q}{ax^2+bx+c}=A\int\frac{2ax+b}{ax^2+bx+c}dx+C\int\frac{1}{ax^2+bx+c}dx\end{array} \)
  • The right-hand side value thus obtained in the previous step can be integrated, and then constant values can be substituted.

4]

\(\begin{array}{l}\int\frac{px+q}{\sqrt{ax^2+bx+c}}\end{array} \)

Tricks to Solve:

In this case, we proceed as follows:

  • The numerator is expressed as the sum of constant times differentiation of the denominator and a constant; that is, px + q is written as A (2ax + b) + C, where 2ax + b=d/dx (ax2+bx+c). A and C are arbitrary constants.
  • The values of constants are computed by equating the coefficients of like powers on both sides.
  • Replace the numerator (px + q) by A (2ax + b) + C in the given integral, which will give,
    \(\begin{array}{l}\int\frac{px+q}{\sqrt{ax^2+bx+c}}dx=A\int\frac{2ax+b}{\sqrt{ax^2+bx+c}}dx+C\int \frac{1}{\sqrt{ax^2+bx+c}}dx\end{array} \)
  • The right-hand side value, thus obtained in the previous step, can be integrated, and then, constant values can be substituted.

5]

\(\begin{array}{l}\int \sqrt{ax^2+bx+c}dx\end{array} \)

Tricks to Solve:

In such types of integrals, the following steps are followed:

  • Make the coefficient of x2 as unity.
  • \(\begin{array}{l}\text{To obtain the form}\ [x+\frac{b}{2a}]^2 + \frac{4ac-b^2}{4a^2},\end{array} \)
    add and subtract the square of half of the coefficient of x.
  • Hence, the integral reduces to,
    \(\begin{array}{l}\int \sqrt{a^2+x^2}dx\ \text{or}\ \int \sqrt{a^2-x^2}dx\ \text{or}\ \int \sqrt{x^2-a^2}dx.\end{array} \)
  • These integrals can be computed easily by using direct formulae.

Also Read

Definite and Indefinite Integration

Integration of Functions

Applications of Integration

Algebraic Integrals Examples

Below are some solved examples on algebraic integrals for IIT JEE aspirants.

Example 1:

\(\begin{array}{l}\int \frac{x+1}{\sqrt{2x^2+x-3}}dx\end{array} \)

Solution:

By completing the square,

Algebraic Integrals Example

Algebraic Integrals Example for IIT JEE

Example 2:

\(\begin{array}{l}\int \frac{1}{x^2+6x+5}dx\end{array} \)

Solution:

Algebraic Expression Problem

By completing the square,

Example on Algebraic Integrals

Example 3:

\(\begin{array}{l}\int \frac{1}{\sqrt{6x^2-5x+1}}dx\end{array} \)

Solution:

Example on Algebraic Integrals for IIT JEE

By completing the square,

Solved Problem on Algebraic Integrals

Solved Problems on Algebraic Integrals

Example 4:

\(\begin{array}{l} I=\int _0^{10}\left|sinx\right|\:dx\\\end{array} \)

Solution:Β 

\(\begin{array}{l}=\int _0^{\pi }\sin \left(x\right)dx+\int _{\pi }^{2\pi }-\sin \left(x\right)dx+\int _{2\pi }^{3\pi }\sin \left(x\right)dx+\int _{3\pi }^{10}-\sin \left(x\right)dx\\\int _0^{\pi }\sin \left(x\right)dx=\left[-\cos \left(x\right)\right]^{\pi }_0=2\\ \int _{\pi }^{2\pi }-\sin \left(x\right)dx=-\int _{\pi }^{2\pi }\sin \left(x\right)dx=-\left[-\cos \left(x\right)\right]^{2\pi }_{\pi }=-\left(-2\right)\\\int _{2\pi }^{3\pi }\sin \left(x\right)dx=\left[-\cos \left(x\right)\right]^{3\pi }_{2\pi }=2\\\int _{3\pi }^{10}-\sin \left(x\right)dx=-\left[-\cos \left(x\right)\right]^{10}_{3\pi }=-\left(-\cos \left(10\right)-1\right)\\=2+2+2+\cos \left(10\right)+1\\=\cos \left(10\right)+7\end{array} \)

Example 5:

\(\begin{array}{l}\text{If}\ I_{n}=\int_{0}^{\pi/4}tan^{n}xdx,\ \text{then}\ \lim_{n\rightarrow \infty}n[I_{n}+I_{n+2}]=\end{array} \)

Solution:

\(\begin{array}{l}I_{n}+I_{n+2}=\int_{0}^{\pi/4}tan^{n}x[1+tan^{2}xdx]\\ \int_{0}^{\pi/4}tan^{n}xsec^{2}xdx=\frac{tan^{n+1}x}{n+1} \text \ from \ 0 \ to \ \frac{\pi}{4}\\ =\frac{1-0}{n+1}=\frac{1}{n+1}\\ I_{n}+I_{n+2}=\frac{1}{n+1}\\\Rightarrow \lim_{x\rightarrow \infty }n[I_{n}+I_{n+2}]\\ \Rightarrow \lim_{x\rightarrow \infty }n*\frac{1}{n+1}=\frac{n}{n+1}=\frac{n}{n*[1+\frac{1}{n}]}=1\end{array} \)

Example 6:

\(\begin{array}{l}\int _0^1\sqrt[3]{1+7x}dx\end{array} \)

Solution:

\(\begin{array}{l}\mathrm{Apply\:u-substitution:}\:u=1+7x\\=\int _1^8\frac{\sqrt[3]{u}}{7}du\\=\frac{1}{7}\cdot \int _1^8\sqrt[3]{u}du\\=\frac{1}{7}\cdot \int _1^8u^{\frac{1}{3}}du\\=\frac{1}{7}\left[\frac{u^{\frac{1}{3}+1}}{\frac{1}{3}+1}\right]^8_1\\=\frac{1\cdot \:\left[\frac{u^{\frac{1}{3}+1}}{\frac{1}{3}+1}\right]^8_1}{7}\\\frac{u^{\frac{1}{3}+1}}{\frac{1}{3}+1}=\frac{3u^{\frac{4}{3}}}{4}\\=\frac{\left[\frac{u^{\frac{4}{3}}\cdot \:3}{4}\right]^8_1}{7}\\=\frac{1}{7}\left[\frac{u^{\frac{4}{3}}\cdot \:3}{4}\right]^8_1\\=\frac{1}{7}\cdot \frac{45}{4}\\=\frac{45}{28}\end{array} \)

Example 7:

\(\begin{array}{l}\int \frac{e^x}{e^x+e^{-x}}dx\end{array} \)

Solution:Β 

\(\begin{array}{l}\int \frac{e^x}{e^x+e^{-x}}dx\\ \mathrm{Apply\:u-substitution:}\:u=e^x\\ =\int \frac{u}{u^2+1}du\\ \mathrm{Apply\:u-substitution:}\:v=u^2+1\\=\int \frac{1}{2v}dv\\\frac{1}{2}\ln \left|v\right|\\=\frac{1}{2}\ln \left|\left(e^x\right)^2+1\right|\\=\frac{1}{2}\ln \left|e^{2x}+1\right|\\=\frac{1}{2}\ln \left|e^{2x}+1\right|+C\end{array} \)

Integrations Important JEE Main Questions

Definite Integration JEE Questions

Indefinite Integration JEE Questions

Frequently Asked Questions

Q1

Name some methods of integration.

Integration by parts, integration by substitution, integration using trigonometric identities, and integration by partial fraction are some methods of integration.

Q2

What is the value of ∫aa f(x) dx?

The value of ∫aa f(x) dx = 0.

Q3

If f(x) is an odd function, what is βˆ«β€“aa f(x) dx?

If f(x) is an odd function, βˆ«β€“aa f(x) dx = 0.

Q4

If f(x) is an even function, what is βˆ«β€“aa f(x) dx?

If f(x) is an even function, βˆ«β€“aa f(x) dx = 2∫0a f(x) dx.

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