JEE Main 2024 Question Paper Solution Discussion Live JEE Main 2024 Question Paper Solution Discussion Live

Integration of Functions

Integration is the process of finding the antiderivative. The integration of g′(x), with respect to dx, is given by

∫ g′(x) dx = g(x) + C, where C is the constant of integration.

The two types of integrals are

  • Definite integral: An integral with limits, namely upper and lower limits, without the constant of integration.
  • Indefinite integral: An integral without limits and with an arbitrary constant.

This article covers standard integrals, properties of integration, important formulas and examples of integration which helps students to have a deep knowledge of the topic.

Standard Integrals

Integrals of Rational and Irrational Functions

\(\begin{array}{l}\int x^n dx = \frac{x^{n+1}}{n+1} + C , n \ne 1\\ \int \frac{1}{x} dx = \ln|x| + C \\ \int c \, dx = c \cdot x + C \\ \int x \, dx = \frac{x^2}{2} + C \\ \int x^2 \, dx = \frac{x^3}{3} + C \\ \int \frac{1}{x^2} dx = -\frac{1}{x} + C \\\end{array} \)
\(\begin{array}{l}\int \sqrt{x} \, dx = \frac{2\cdot x \cdot \sqrt{x} }{3} + C\\ \int \frac{1}{1+x^2} dx = \arctan x + C \\ \int \frac{1}{\sqrt{1-x^2}} dx = \arcsin x + C \\\end{array} \)

Integrals of Trigonometric Functions

\(\begin{array}{l}\int \sin x\,dx = -\cos x + C \\ \int \cos x\,dx = \sin x + C\\ \int \tan x\,dx = \ln|\sec x| + C \\ \int \sec x\,dx = \ln|\tan x + \sec x | + C \\\end{array} \)
\(\begin{array}{l}\int \sin^2x\,dx = \frac{1}{2}(x-\sin x \cdot \cos x) + C\\ \int \cos^2x\,dx = \frac{1}{2}(x + \sin x \cdot \cos x) + C \\ \int \tan^2x\,dx = \tan x – x + C \\ \int \sec^2x\,dx = \tan x + C \\\end{array} \)

Integrals of Exponential and Logarithmic Functions

\(\begin{array}{l}\int \ln x \,dx =x \cdot \ln x -x + C \\ \int x^n \cdot \ln x \,dx =\frac{x^{n+1}}{n+1} \ln x – \frac{x^{n+1}}{(n+1)^2} + C \\ \int e^x\,dx = e^x + C \\ \int a^x\,dx = \frac{a^x}{\ln a} + C\\\end{array} \)

Properties of  Integration

Property 1:

\(\begin{array}{l}\int\limits_{a}^{a}{f(x)\,dx=0}\end{array} \)

Property 2:

\(\begin{array}{l}\int\limits_{a}^{b}{f(x)dx=}-\int\limits_{b}^{a}{f(x)dx}\end{array} \)

Property 3:

\(\begin{array}{l}\int\limits_{a}^{b}{f(x)dx=}\int\limits_{a}^{b}{f(t)dt}\end{array} \)

Property 4:

\(\begin{array}{l}\int\limits_{a}^{b}{f(x)dx=}\int\limits_{a}^{c}{f(x)dx+\int\limits_{a}^{b}{f(x)dx}}\end{array} \)

Property 5:

\(\begin{array}{l}(i)\ \int\limits_{a}^{b}{f(x)dx=}\int\limits_{a}^{b}{f(a+b-x)dx}\end{array} \)
\(\begin{array}{l}(ii)\ \int\limits_{0}^{a}{f(x)dx=}\int\limits_{0}^{a}{f(a-x)dx}\end{array} \)

 

⇒ Also Read: Definite and Indefinite Integration

Useful Formulas

  • \(\begin{array}{l}\int{{{e}^{ax}}\sin bx=\frac{{{e}^{ax}}}{{{a}^{2}}+{{b}^{2}}}\left[ a\sin bx-b\cos bx \right]}\end{array} \)
  • \(\begin{array}{l}\int{{{e}^{ax}}\cos bx=\frac{{{e}^{ax}}}{{{a}^{2}}+{{b}^{2}}}\left[ a\cos bx+b\sin bx \right]}\end{array} \)
  • \(\begin{array}{l}\int{{{e}^{x}}\left( f(x)+f'(x) \right)}={{e}^{x}}f(x)\end{array} \)

Illustration:

\(\begin{array}{l}\int{{{e}^{x}}(\sin x+\cos x)dx={{e}^{x}}\sin x+c\\}\end{array} \)

 

\(\begin{array}{l}\int{{{e}^{x}}(lnx+\frac{1}{x})dx={{e}^{x}}lnx+c}\end{array} \)

Integration of Trigonometric Functions

Type 1:

\(\begin{array}{l}I=\int{{{\sin }^{m}}x{{\cos }^{n}}xdx}\end{array} \)

1. If m is odd, put cos x = t

2. If n is odd, put sin x = t

3. If m, n rationales then put tan x = t

4. If both are even, then use the reduction method.

\(\begin{array}{l}Q\int{\frac{{{\cos }^{3}}x}{{{\sin }^{6}}x}dx=\int{\frac{1-{{t}^{2}}}{{{t}^{6}}}dt}}\end{array} \)

Where t = sin x

\(\begin{array}{l}=\int{{{t}^{-6}}-{{t}^{-4}}dt}\end{array} \)
\(\begin{array}{l}=-\frac{1}{5si{{n}^{5}}x}+\frac{1}{3{{\sin }^{3}}x}+c\end{array} \)

Type 2:

\(\begin{array}{l}\int{\frac{dx}{a\cos x+b\sin x+c}}\end{array} \)

Put t = tan (x/2)

Illustration

\(\begin{array}{l}\int{\frac{dx}{2+\sin x}}\end{array} \)
\(\begin{array}{l}\Rightarrow t=\tan \left( \frac{x}{2} \right)\end{array} \)
\(\begin{array}{l}dx=\frac{2dt}{1+{{t}^{2}}}\end{array} \)
\(\begin{array}{l}=\int{\frac{\frac{2dt}{1+{{t}^{2}}}}{2+\frac{2t}{1+{{t}^{2}}}}}\end{array} \)
\(\begin{array}{l}\Rightarrow \int{\frac{dt}{{{t}^{2}}+t+1}}\end{array} \)
\(\begin{array}{l}=\frac{2}{\sqrt{3}}{{\tan }^{-1}}\left( \frac{2t+1}{\sqrt{3}} \right)\end{array} \)
\(\begin{array}{l}=\frac{2}{\sqrt{3}}{{\tan }^{-1}}\left( \frac{2\tan \frac{x}{2}+1}{\sqrt{3}} \right)+c\end{array} \)

Some Useful Substitutions for Irrational Functions

  • Form 1 :
    \(\begin{array}{l}\int{linear\sqrt{Quadratic}\,dx}\end{array} \)
\(\begin{array}{l}\text{Substitute}\ linear=mQudratic’+n\end{array} \)
  • Form 2:
    \(\begin{array}{l}\int{\frac{dx}{lin\sqrt{li{{n}_{1}}}},}\,\int{\frac{lin}{\sqrt{li{{n}_{1}}}}}dx,\,\int{\frac{\sqrt{li{{n}_{1}}}}{lin}dx}\end{array} \)
\(\begin{array}{l}\text{Substitute}\ li{{n}_{1}}={{t}^{2}}\end{array} \)
  • Form 3:
    \(\begin{array}{l}\int{\frac{1}{lin\sqrt{Qua}}dx}\end{array} \)

Substitute lin = 1/t

  • Form 4:
    \(\begin{array}{l}\int{\frac{dx}{\left( a{{x}^{2}}+b \right)\sqrt{\left( {{x}^{2}}+d \right)}}}\end{array} \)

Substitute x = 1/t and then u2 for at2 + b

Integration Formulas

  1. \(\begin{array}{l}\int\limits_{a}^{b}{f(x)dx=}\int\limits_{a}^{b}{f(t)dt}\end{array} \)
  2. \(\begin{array}{l}\int\limits_{a}^{b}{f(x)dx=}-\int\limits_{b}^{a}{f(x)dx}\end{array} \)
  3. \(\begin{array}{l}\int\limits_{a}^{b}{f(x)dx=}\int\limits_{a}^{c}{f(x)dx}+\int\limits_{c}^{b}{f(x)dx}\end{array} \)
  4. \(\begin{array}{l}\int\limits_{a}^{b}{f(x)dx=}\int\limits_{a}^{b}{f(a+b-x)dx}\end{array} \)
  5. \(\begin{array}{l}\int\limits_{0}^{2a}{f(x)dx}=\int\limits_{0}^{a}{f(x)dx}+\int\limits_{0}^{a}{f(2a-x)dx}\end{array} \)
    \(\begin{array}{l}=0\ \text{if}\\,f(2a-x)=-f(x)\end{array} \)
    and
    \(\begin{array}{l}=2\int\limits_{0}^{a}{f(x)}\ \text{if}\ \,f(2a-x)=f(x)\end{array} \)
  6. \(\begin{array}{l}\int\limits_{-a}^{a}{f(x)dx=\left\{ \begin{matrix} 0 & if\,\,f\left( x \right)\ \text{is odd} \\ 2\int\limits_{0}^{a}{f\left( x \right)dx} & if\,\,f\left( x \right)\ \text{is even} \\ \end{matrix} \right.\,\,\,\,\,\,}\end{array} \)

Problems on Integration

Illustration:

\(\begin{array}{l}\int\limits_{0}^{2}{{{x}^{2}}\left[ x \right]dx=}\int\limits_{0}^{1}{{{x}^{2}}\left[ x \right]dx}+\int\limits_{1}^{2}{{{x}^{2}}\left[ x \right]dx}\end{array} \)
\(\begin{array}{l}=\int\limits_{0}^{1}{{{x}^{2}}.0\,dx}+\int\limits_{1}^{2}{{{x}^{2}}\left[ 1 \right]dx}\end{array} \)
\(\begin{array}{l}=0+\left. \frac{{{x}^{3}}}{3} \right|_{1}^{2}\end{array} \)
\(\begin{array}{l}=\frac{8-1}{3}\,=\frac{7}{3}\end{array} \)

Illustration:

\(\begin{array}{l}\int\limits_{{\pi }/{6}\;}^{{\pi }/{3}\;}{\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx}\end{array} \)
\(\begin{array}{l}I = \int{\frac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx}\ \(\text{by}\ x\to \frac{\pi}{2} – x)\end{array} \)
\(\begin{array}{l}2I=\int\limits_{{\pi }/{6}\;}^{{\pi }/{3}\;}{\frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}}=\int\limits_{{\pi }/{6}\;}^{{\pi }/{3}\;}{1\,dx}=\frac{\pi }{6}\end{array} \)
\(\begin{array}{l}I=\frac{\pi }{12}\end{array} \)

Illustration:

\(\begin{array}{l}I=\int{{{\sin }^{100}}x{{\cos }^{99}}x}\end{array} \)

Here, f(2π – x) = f(x)

Or

\(\begin{array}{l}I=2\int\limits_{0}^{\pi }{{{\sin }^{100}}x{{\cos }^{99}}x}\end{array} \)
\(\begin{array}{l}=2\int\limits_{0}^{\pi }{{{\sin }^{100}}\left( \pi -x \right){{\cos }^{99}}\left( \pi -x \right)}\end{array} \)

I = -I

I = 0

Illustration:

\(\begin{array}{l}\int\limits_{-5}^{5}{{{x}^{3}}=0}\ \text{as}\ f(x)\ \text{is odd}\end{array} \)

Leibnitz’s rule

\(\begin{array}{l}\frac{d}{dx}\int\limits_{u\left( x \right)}^{v\left( x \right)}{f\left( t \right)dt=f\left( v\left( x \right) \right)}\frac{dv\left( x \right)}{dx}-f\left( u\left( x \right) \right)u’\left( x \right)\end{array} \)

Practice Problems

Problem 1.

\(\begin{array}{l}\text{If}\ \int\limits_{{{x}^{2}}}^{{{x}^{3}}}{\frac{1}{\log t}dt=y}\ \text{find}\end{array} \)
\(\begin{array}{l}\frac{dy}{dx}=x\left( x-1 \right){{\left( \log x \right)}^{-1}}\end{array} \)

Problem 2.

\(\begin{array}{l}\text{If}\ \int\limits_{\sin x}^{1}{{{t}^{2}}f\left( t \right)dt=1-\sin x.}\end{array} \)
where x ∈ (0, π/2), find f(1/√3).

Problem 3. 

\(\begin{array}{l}\text{If}\ f\left( 2 \right)=6,f’\left( 2 \right)=\frac{1}{48}.\ \text{Find}\underset{x\to 2}{\mathop{\lim }}\,\int\limits_{6}^{f\left( x \right)}{\frac{4{{t}^{3}}}{x-2}}dt=18.\end{array} \)

Problem 4. 

\(\begin{array}{l}\underset{x\to \infty }{\mathop{\lim }}\,\frac{\int\limits_{0}^{x}{{{e}^{{{x}^{2}}}}dx}}{\int\limits_{0}^{x}{{{e}^{2{{x}^{2}}}}dx}}=0\end{array} \)

Integration by Parts

\(\begin{array}{l}\int{uv\,dx}=u\int{vdx}-\int{u’\left( \int{vdx} \right)}\,dx\end{array} \)

Illustration:

Q.

\(\begin{array}{l}\int{\ell n\,x}=\int{\ell n\,\,x.1\,\,dx}\end{array} \)
\(\begin{array}{l}=x\,\ell n\,x-\int{\frac{1}{x}.\,x}\,dx\end{array} \)
\(\begin{array}{l}=x\,\ell n\,x-x\end{array} \)

Q.

\(\begin{array}{l}\int{x\,{{e}^{x}}dx=x\int{{{e}^{x}}dx-\int{{{\left( 1 \right)}}\left( \int{{{e}^{x}}dx} \right)}dx}}\end{array} \)
\(\begin{array}{l}=x{{e}^{x}}-\int{{{e}^{x}}dx}\end{array} \)
\(\begin{array}{l}=x{{e}^{x}}-{{e}^{x}}\end{array} \)

Integration of Irrational Algebraic Functions

Type

\(\begin{array}{l}\int{\frac{dx}{{{\left( ax+b \right)}^{k}}\sqrt{px+q}}}\end{array} \)

Q.

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

Put x + 1 = t2, we get

\(\begin{array}{l}I=\int{\frac{\left( {{t}^{2}}-1 \right)2t\,dt}{\left( {{t}^{2}}-4 \right)t}}=2\int{\frac{{{t}^{2}}-1}{{{t}^{2}}-4}dt}\end{array} \)
\(\begin{array}{l}=2\int{1}+\frac{3}{{{t}^{2}}-4}dt\end{array} \)
\(\begin{array}{l}=2t+\frac{3}{2}\ell n\left| \frac{t-2}{t+2} \right|+c\end{array} \)
\(\begin{array}{l}=2\sqrt{x+1}+\frac{3}{2}\ell n\left| \frac{\sqrt{x+1}-2}{\sqrt{x+1}+2} \right|+c\end{array} \)

\(\begin{array}{l}\Rightarrow \int\limits_{0}^{2a}{f\left( x \right)dx=\int\limits_{0}^{a}{f\left( x \right)dx+}}\int\limits_{0}^{a}{f\left( 2a-x \right)dx}\end{array} \)

= 0 if f(2a – x) = -f(x)

\(\begin{array}{l}=2\int\limits_{0}^{a}{f\left( x \right)}\ \text{if}\ f(2a – x) = f(x)\end{array} \)

Optimisation of Area for Greatest and Least Values

Illustration:

If area by y = f(x) and y = x2 + 2 between abscissa x = 2 and x = α is α3 – 4α2 + 8. Find f(x). 

Answer:

\(\begin{array}{l}{{\alpha }^{3}}-4{{\alpha }^{2}}+8=\int\limits_{2}^{\alpha }{\left( {{x}^{2}}+2-f\left( x \right) \right)dx}\end{array} \)

Differentiating with Labniz equation,

\(\begin{array}{l}3{{\alpha }^{2}}-8\alpha ={{\alpha }^{2}}+2-f\left( \alpha \right)\end{array} \)
\(\begin{array}{l}f\left( x \right)=-2{{x}^{2}}+8x+2\end{array} \)

Integrations Important JEE Main Questions

Definite Integration JEE Questions

Indefinite Integration JEE Questions

 

Solved Problems on Integration

Problem 1:

\(\begin{array}{l}\int_{{}}^{{}}{\frac{dx}{\cos (x-a)\cos (x-b)}=}\\\end{array} \)

Solution:

\(\begin{array}{l}\int_{{}}^{{}}{\frac{dx}{\cos (x-a)\cos (x-b)}}\\ =\frac{1}{\sin (a-b)}\int_{{}}^{{}}{\frac{\sin \left\{ (x-b)-(x-a) \right\}}{\cos (x-a)\,.\,\cos (x-b)}\,dx}\\ =\frac{1}{\sin (a-b)}\int_{{}}^{{}}{\left\{ \frac{\sin (x-b)}{\cos (x-b)}-\frac{\sin (x-a)}{\cos (x-a)} \right\}dx}\\ =\text{cosec}\,(a-b)\log \frac{\cos (x-a)}{\cos (x-b)}+c\end{array} \)

Problem 2:

\(\begin{array}{l}\int_{{}}^{{}}{\frac{dx}{\sqrt{x+a}+\sqrt{x+b}}}=\\\end{array} \)

Solution:

\(\begin{array}{l}\int_{{}}^{{}}{\frac{dx}{\sqrt{x+a}+\sqrt{x+b}}=\int_{{}}^{{}}{\frac{\sqrt{x+a}-\sqrt{x+b}}{(x+a)-(x+b)}\,dx}}\\ =\frac{1}{(a-b)}\int_{{}}^{{}}{{{(x+a)}^{1/2}}dx}-\frac{1}{(a-b)}\int_{{}}^{{}}{{{(x+b)}^{1/2}}dx}\\ =\frac{2}{3(a-b)}[{{(x+a)}^{3/2}}-{{(x+b)}^{3/2}}]+c\end{array} \)

Problem 3:

\(\begin{array}{l}\int_{{}}^{{}}{\frac{{{x}^{3}}-x-2}{(1-{{x}^{2}})}\ dx=}\\\end{array} \)

Solution:

\(\begin{array}{l}\int_{{}}^{{}}{\frac{{{x}^{3}}-x-2}{(1-{{x}^{2}})}\,dx}\\=\int_{{}}^{{}}{\frac{-x(1-{{x}^{2}})}{(1-{{x}^{2}})}\,dx-\int_{{}}^{{}}{\frac{2}{1-{{x}^{2}}}\,dx}}\\=-\int_{{}}^{{}}{x\,dx}-2\int_{{}}^{{}}{\frac{1}{1-{{x}^{2}}}\,dx\\=\frac{-{{x}^{2}}}{2}+\log \left(\frac{x-1}{x+1}\right)+c.}\end{array} \)

Problem 4:

\(\begin{array}{l}\int_{{}}^{{}}{\frac{{{\sin }^{8}}x-{{\cos }^{8}}x}{1-2{{\sin }^{2}}x{{\cos }^{2}}x}\ dx=}\\\end{array} \)

Solution:

\(\begin{array}{l}\int_{{}}^{{}}{\frac{{{\sin }^{8}}x-{{\cos }^{8}}x}{1-2{{\sin }^{2}}x{{\cos }^{2}}x}\,dx}\\ =\int_{{}}^{{}}{\frac{({{\sin }^{4}}x+{{\cos }^{4}}x)({{\sin }^{4}}x-{{\cos }^{4}}x)}{{{({{\sin }^{2}}x+{{\cos }^{2}}x)}^{2}}-2{{\sin }^{2}}x{{\cos }^{2}}x}}\,dx\\ =\int_{{}}^{{}}{({{\sin }^{4}}x-{{\cos }^{4}}x)\,dx}\\ =\int_{{}}^{{}}{({{\sin }^{2}}x+{{\cos }^{2}}x)({{\sin }^{2}}x-{{\cos }^{2}}x)\,dx}\\ =\int_{{}}^{{}}{({{\sin }^{2}}x-{{\cos }^{2}}x)\,dx}\\=\int_{{}}^{{}}{-\cos 2x\,dx=-\frac{\sin 2x}{2}+c}\end{array} \)

Problem 5:

\(\begin{array}{l}\int_{{}}^{{}}{\frac{{{x}^{2}}dx}{{{(a+bx)}^{2}}}}=\\\end{array} \)

Solution:
Put a + bx = t ⇒ x = (t – a)/b and dx = dt/b

\(\begin{array}{l}I={{\int_{{}}^{{}}{\left( \frac{t-a}{b} \right)}}^{2}}\times \frac{1}{{{t}^{2}}}\frac{dt}{b}\\ =\frac{1}{{{b}^{2}}}\int_{{}}^{{}}{\left( 1-\frac{2a}{t}+{{a}^{2}}.{{t}^{-2}} \right)}\,dt\\=\frac{1}{{{b}^{2}}}\left[ t-2a\,\,\log t-\frac{{{a}^{2}}}{t} \right]\\ =\frac{1}{{{b}^{2}}}\left[ x+\frac{a}{b}-\frac{2a}{b}\log (a+bx)-\frac{{{a}^{2}}}{b}\frac{1}{(a+bx)} \right]\\\end{array} \)

Problem 6: Solve

\(\begin{array}{l}\int{\frac{2\cos x+3\sin x}{4\cos x+5\sin x}}\,dx\end{array} \)

Solution:

Problem of type

\(\begin{array}{l}\int{\frac{a\cos x+b\sin x+p}{c\cos x+d\sin x+q}}\,dx,\int{\frac{a{{e}^{x}}+b{{e}^{-x}}+c}{d{{e}^{x}}+f{{e}^{-x}}+h}}\,dx\end{array} \)
can be solved by
\(\begin{array}{l}Nr=nDr+mDr'\end{array} \)

Now,

Let 2 cos x + 3 sin x = a( 4 cos x + 5 sin x) + b(-4 sin x + 5 cos x)

Solving by comparing, we get

\(\begin{array}{l}a=\frac{23}{41}\,\,\,\,\,\,b=\frac{-2}{41}\end{array} \)
\(\begin{array}{l}\therefore I=\int{\frac{23}{41}-\frac{2}{41}\left( \frac{-4\sin x+5\cos x}{4\cos x+5\sin x} \right)dx}\end{array} \)
\(\begin{array}{l}=\frac{23}{41}x-\frac{2}{41}\ell n\left| 4\cos x+5\sin x \right|+c\end{array} \)

Problem 7:  Find area between y2 ≤ 4x, x2 + y2 ≥ 2x and x ≤ y + 2 in first quadrant.

Answer:

Problems on integration example 4

\(\begin{array}{l}A=\int\limits_{0}^{{{\left( \sqrt{3}+1 \right)}^{2}}}{\sqrt{4x}\,\,dx-} ar(semicircle) – ar(\triangle ABC)\end{array} \)
\(\begin{array}{l}=\sqrt{4}\left[ \frac{2{{x}^{3/2}}}{3} \right]_{0}^{{{\left( \sqrt{3}+1 \right)}^{2}}}-\frac{\pi }{2}-\frac{1}{2}{{\left( \sqrt{3}+1 \right)}^{2}}2\left( \sqrt{3}+1 \right)\end{array} \)
\(\begin{array}{l}=\frac{{{\left( \sqrt{3}+1 \right)}^{3}}}{3}-\frac{\pi }{2}\end{array} \)

Most Important Questions from Definite Integration for JEE Advanced

Frequently Asked Questions

Q1

What do you mean by integration in maths?

Integration is the process of finding the antiderivative of a function.

Q2

What is the integral of x?

The integral of x = (x2/2) + C, where C is the constant of integration.

Q3

What is the integral of sin x?

Integral of sin x =-cos x + C

Q4

Give two applications of integration.

Integration is used to find the area under a curve. It is also used to find the velocity and trajectory of a satellite.

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