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 include:
Definite Integral: An integral with limits namely upper and lower limit 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 on integration which helps students to have a deep knowledge on the topic.
Standard Integrals
Integrals of Rational and Irrational Functions
∫ x n d x = x n + 1 n + 1 + C , n ≠ 1 ∫ 1 x d x = ln ∣ x ∣ + C ∫ c d x = c ⋅ x + C ∫ x d x = x 2 2 + C ∫ x 2 d x = x 3 3 + C ∫ 1 x 2 d x = − 1 x + C \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 \\ ∫ x n d x = n + 1 x n + 1 + C , n = 1 ∫ x 1 d x = ln ∣ x ∣ + C ∫ c d x = c ⋅ x + C ∫ x d x = 2 x 2 + C ∫ x 2 d x = 3 x 3 + C ∫ x 2 1 d x = − x 1 + C ∫ x d x = 2 ⋅ x ⋅ x 3 + C ∫ 1 1 + x 2 d x = arctan x + C ∫ 1 1 − x 2 d x = arcsin x + C \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 \\ ∫ x d x = 3 2 ⋅ x ⋅ x + C ∫ 1 + x 2 1 d x = arctan x + C ∫ 1 − x 2 1 d x = arcsin x + C Integrals of Trigonometric Functions
∫ sin x d x = − cos x + C ∫ cos x d x = sin x + C ∫ tan x d x = ln ∣ sec x ∣ + C ∫ sec x d x = ln ∣ tan x + sec x ∣ + C \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 \\ ∫ sin x d x = − cos x + C ∫ cos x d x = sin x + C ∫ tan x d x = ln ∣ sec x ∣ + C ∫ sec x d x = ln ∣ tan x + sec x ∣ + C ∫ sin 2 x d x = 1 2 ( x − sin x ⋅ cos x ) + C ∫ cos 2 x d x = 1 2 ( x + sin x ⋅ cos x ) + C ∫ tan 2 x d x = tan x – x + C ∫ sec 2 x d x = tan x + C \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 \\ ∫ sin 2 x d x = 2 1 ( x − sin x ⋅ cos x ) + C ∫ cos 2 x d x = 2 1 ( x + sin x ⋅ cos x ) + C ∫ tan 2 x d x = tan x – x + C ∫ sec 2 x d x = tan x + C Integrals of Exponential and Logarithmic Functions
∫ ln x d x = x ⋅ ln x − x + C ∫ x n ⋅ ln x d x = x n + 1 n + 1 ln x – x n + 1 ( n + 1 ) 2 + C ∫ e x d x = e x + C ∫ a x d x = a x ln a + C \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\\ ∫ ln x d x = x ⋅ ln x − x + C ∫ x n ⋅ ln x d x = n + 1 x n + 1 ln x – ( n + 1 ) 2 x n + 1 + C ∫ e x d x = e x + C ∫ a x d x = l n a a x + C Properties of Integration Property 1: ∫ a a f ( x ) d x = 0 \int\limits_{a}^{a}{f(x)\,dx=0} a ∫ a f ( x ) d x = 0
Property 2: ∫ a b f ( x ) = − ∫ b a f ( x ) d x \int\limits_{a}^{b}{f(x)=}-\int\limits_{b}^{a}{f(x)dx} a ∫ b f ( x ) = − b ∫ a f ( x ) d x
Property 3: ∫ a b f ( x ) = ∫ a b f ( t ) \int\limits_{a}^{b}{f(x)=}\int\limits_{a}^{b}{f(t)} a ∫ b f ( x ) = a ∫ b f ( t )
Property 4: ∫ a b f ( x ) d x = ∫ a c f ( x ) d x + ∫ a b f ( x ) d x \int\limits_{a}^{b}{f(x)dx=}\int\limits_{a}^{c}{f(x)dx+\int\limits_{a}^{b}{f(x)dx}} a ∫ b f ( x ) d x = a ∫ c f ( x ) d x + a ∫ b f ( x ) d x
Property 5: (i) ∫ a b f ( x ) d x = ∫ a b f ( a + b − x ) d x \int\limits_{a}^{b}{f(x)dx=}\int\limits_{a}^{b}{f(a+b-x)dx} a ∫ b f ( x ) d x = a ∫ b f ( a + b − x ) d x
(ii) ∫ 0 a f ( x ) d x = ∫ 0 a f ( a − x ) d x \int\limits_{0}^{a}{f(x)dx=}\int\limits_{0}^{a}{f(a-x)dx} 0 ∫ a f ( x ) d x = 0 ∫ a f ( a − x ) d x
(iii) f ( x ) = { x 2 + x 1 ≤ x ≤ 2 3 x 2 ≤ x ≤ 3 f(x)=\left\{ \begin{matrix} {{x}^{2}}+x & 1\le x\le 2 \\ 3x & 2\le x\le 3 \\ \end{matrix} \right. f ( x ) = { x 2 + x 3 x 1 ≤ x ≤ 2 2 ≤ x ≤ 3 ∫ 1 3 f ( x ) = 34 3 \int\limits_{1}^{3}{f(x)=\frac{34}{3}} 1 ∫ 3 f ( x ) = 3 3 4
(iv) ∫ − 1 1 e ( x ) d x = 2 ( e − 1 ) \int\limits_{-1}^{1}{{{e}^{(x)}}}dx=2(e-1) − 1 ∫ 1 e ( x ) d x = 2 ( e − 1 )
(v) ∫ − 2 3 ∣ x ∣ d x = 13 2 \int\limits_{-2}^{3}{\left| \left. x \right| \right.dx=\frac{13}{2}} − 2 ∫ 3 ∣ x ∣ d x = 2 1 3 ∫ − 1 1 x 3 ∣ x ∣ d x = 0 \int\limits_{-1}^{1}{{{x}^{3}}\left| x \right|dx=0} − 1 ∫ 1 x 3 ∣ x ∣ d x = 0
(vi) ∫ 2 8 ∣ x − 5 ∣ d x = 9 \int\limits_{2}^{8}{\left| x-5 \right|dx=9} 2 ∫ 8 ∣ x − 5 ∣ d x = 9 ∫ 1 / 4 1 ∣ 2 x − 1 ∣ d x = 5 16 \int\limits_{{}^{1}/{}_{4}}^{1}{\left| 2x-1 \right|dx=\frac{5}{16}} 1 / 4 ∫ 1 ∣ 2 x − 1 ∣ d x = 1 6 5
(vii) ∫ π / 2 π / 2 ∣ sin x ∣ d x = 2 \int\limits_{{}^{\pi }/{}_{2}}^{{}^{\pi }/{}_{2}}{\left| \sin x \right|dx=2} π / 2 ∫ π / 2 ∣ sin x ∣ d x = 2 ∫ − 2 2 ∣ x ∣ + ∣ x − 1 ∣ d x = 9 \int\limits_{-2}^{2}{\left| x \right|+\left| x-1 \right|dx=9} − 2 ∫ 2 ∣ x ∣ + ∣ x − 1 ∣ d x = 9
⇒ Also Read Definite and Indefinite Integration
Useful Formulas
∫ e a x sin b x = e a x a 2 + b 2 [ a sin b x − b cos b x ] \int{{{e}^{ax}}\sin bx=\frac{{{e}^{ax}}}{{{a}^{2}}+{{b}^{2}}}\left[ a\sin bx-b\cos bx \right]} ∫ e a x sin b x = a 2 + b 2 e a x [ a sin b x − b cos b x ] ∫ e a x cos b x = e a x a 2 + b 2 [ a cos b x + b sin b x ] \int{{{e}^{ax}}\cos bx=\frac{{{e}^{ax}}}{{{a}^{2}}+{{b}^{2}}}\left[ a\cos bx+b\sin bx \right]} ∫ e a x cos b x = a 2 + b 2 e a x [ a cos b x + b sin b x ]
Illustration:
∫ e − 2 x sin x cos 3 x d x \int{{{e}^{-2x}}\sin x\cos 3xdx} ∫ e − 2 x sin x cos 3 x d x = ∫ e − 2 x ( sin 4 x − sin 2 x ) d x =\int{{{e}^{-2x}}\left( \sin 4x-\sin 2x \right)dx} = ∫ e − 2 x ( sin 4 x − sin 2 x ) d x = e − 2 x [ ( sin 2 x + cos 2 x 8 ) − sin 4 x + 2 cos 4 x 20 ] ={{e}^{-2x}}\left[ \left( \frac{\sin 2x+\cos 2x}{8} \right)-\frac{\sin 4x+2\cos 4x}{20} \right] = e − 2 x [ ( 8 s i n 2 x + c o s 2 x ) − 2 0 s i n 4 x + 2 c o s 4 x ] ∫ e x ( f ( x ) + f ′ ( x ) ) = e x f ( x ) \int{{{e}^{x}}\left( f(x)+f'(x) \right)}={{e}^{x}}f(x) ∫ e x ( f ( x ) + f ′ ( x ) ) = e x f ( x )
Illustration:
∫ e x ( 1 x + 1 x 2 ) d x = e x x + c \int{{{e}^{x}}\left( \frac{1}{x}+\frac{1}{{{x}^{2}}} \right)dx}={{\frac{e}{x}}^{x}}+c ∫ e x ( x 1 + x 2 1 ) d x = x e x + c ∫ e x ( sin x + cos x ) d x = e x sin x + c \int{{{e}^{x}}(\sin x+\cos x)dx={{e}^{x}}\sin x+c} ∫ e x ( sin x + cos x ) d x = e x sin x + c ∫ e x ( l n x + 1 x ) d x = e x l n x + c \int{{{e}^{x}}(lnx+\frac{1}{x})dx={{e}^{x}}lnx+c} ∫ e x ( l n x + x 1 ) d x = e x l n x + c Integration of Trigonometric Functions Type 1: I = ∫ sin m x cos n x d x I=\int{{{\sin }^{m}}x{{\cos }^{n}}xdx} I = ∫ sin m x cos n x d x
1. If m –odd put cos x = t \cos x=t cos x = t
2. If n odd put sin x = t \sin x=t sin x = t
3. If m, n rationales then put tan x = t \tan x=t tan x = t
4. If both even then use reduction method
Q ∫ cos 3 x sin 6 x d x = ∫ 1 − t 2 t 6 d t Q\int{\frac{{{\cos }^{3}}x}{{{\sin }^{6}}x}dx=\int{\frac{1-{{t}^{2}}}{{{t}^{6}}}dt}} Q ∫ s i n 6 x c o s 3 x d x = ∫ t 6 1 − t 2 d t Where t = sin x t=\sin x t = sin x = ∫ t − 6 − t − 4 d t =\int{{{t}^{-6}}-{{t}^{-4}}dt} = ∫ t − 6 − t − 4 d t = − 1 5 s i n 5 x + 1 3 sin 3 x + c =-\frac{1}{5si{{n}^{5}}x}+\frac{1}{3{{\sin }^{3}}x}+c = − 5 s i n 5 x 1 + 3 s i n 3 x 1 + c
Type 2: ∫ d x a cos x + b sin x + c \int{\frac{dx}{a\cos x+b\sin x+c}} ∫ a c o s x + b s i n x + c d x
Put t = tan ( x 2 ) t=\tan \left( \frac{x}{2} \right) t = tan ( 2 x )
Illustration
∫ d x 2 + sin x \int{\frac{dx}{2+\sin x}} ∫ 2 + s i n x d x ⇒ t = tan ( x 2 ) \Rightarrow t=\tan \left( \frac{x}{2} \right) ⇒ t = tan ( 2 x ) d x = 2 d t 1 + t 2 dx=\frac{2dt}{1+{{t}^{2}}} d x = 1 + t 2 2 d t = ∫ 2 d t 1 + t 2 2 + 2 t 1 + t 2 =\int{\frac{\frac{2dt}{1+{{t}^{2}}}}{2+\frac{2t}{1+{{t}^{2}}}}} = ∫ 2 + 1 + t 2 2 t 1 + t 2 2 d t ⇒ ∫ d t t 2 + t + 1 \Rightarrow \int{\frac{dt}{{{t}^{2}}+t+1}} ⇒ ∫ t 2 + t + 1 d t = 2 3 tan − 1 ( 2 t + 1 3 ) =\frac{2}{\sqrt{3}}{{\tan }^{-1}}\left( \frac{2t+1}{\sqrt{3}} \right) = 3 2 tan − 1 ( 3 2 t + 1 ) = 2 3 tan − 1 ( 2 tan x 2 + 1 3 ) + c =\frac{2}{\sqrt{3}}{{\tan }^{-1}}\left( \frac{2\tan \frac{x}{2}+1}{\sqrt{3}} \right)+c = 3 2 tan − 1 ( 3 2 t a n 2 x + 1 ) + c Some useful Substitutions for Irrational Functions
Form 1 : ∫ l i n e a r Q u a d r a l i s d x \int{linear\sqrt{Quadralis}\,dx} ∫ l i n e a r Q u a d r a l i s d x
Substitute l i n e r = m Q u d r a t i c ’ + n liner=mQudratic’+n l i n e r = m Q u d r a t i c ’ + n
Form 2: ∫ d x l i n l i n 1 , ∫ l i n l i n 1 d x , ∫ l i n 1 l i n d x \int{\frac{dx}{lin\sqrt{li{{n}_{1}}}},}\,\int{\frac{lin}{\sqrt{li{{n}_{1}}}}}dx,\,\int{\frac{\sqrt{li{{n}_{1}}}}{lin}dx} ∫ l i n l i n 1 d x , ∫ l i n 1 l i n d x , ∫ l i n l i n 1 d x
Substitute l i n 1 = t 2 li{{n}_{1}}={{t}^{2}} l i n 1 = t 2
Form 3: ∫ 1 l i n Q u a d x \int{\frac{1}{lin\sqrt{Qua}}dx} ∫ l i n Q u a 1 d x
Substitute l i n = 1 t lin=\frac{1}{t} l i n = t 1
Form 4: ∫ d x ( a x 2 + b ) ( x 2 + d ) \int{\frac{dx}{\left( a{{x}^{2}}+b \right)\sqrt{\left( {{x}^{2}}+d \right)}}} ∫ ( a x 2 + b ) ( x 2 + d ) d x
Substitute x = 1 t x=\frac{1}{t} x = t 1 u 2 {{u}^{2}} u 2 a t 2 + b a{{t}^{2}}+b a t 2 + b
Integration Formulas
∫ a b f ( x ) d x = ∫ a b f ( t ) d t \int\limits_{a}^{b}{f(x)dx=}\int\limits_{a}^{b}{f(t)dt} a ∫ b f ( x ) d x = a ∫ b f ( t ) d t ∫ a b f ( x ) d x = − ∫ b a f ( x ) d x \int\limits_{a}^{b}{f(x)dx=}-\int\limits_{b}^{a}{f(x)dx} a ∫ b f ( x ) d x = − b ∫ a f ( x ) d x ∫ a b f ( x ) d x = ∫ a c f ( x ) d x + ∫ c b f ( x ) d x \int\limits_{a}^{b}{f(x)dx=}\int\limits_{a}^{c}{f(x)dx}+\int\limits_{c}^{b}{f(x)dx} a ∫ b f ( x ) d x = a ∫ c f ( x ) d x + c ∫ b f ( x ) d x ∫ a b f ( x ) d x = ∫ a b f ( a + b − x ) d x \int\limits_{a}^{b}{f(x)dx=}\int\limits_{a}^{b}{f(a+b-x)dx} a ∫ b f ( x ) d x = a ∫ b f ( a + b − x ) d x ∫ 0 2 a f ( x ) d x = ∫ 0 a f ( x ) d x + ∫ 0 a f ( 2 a − x ) d x \int\limits_{0}^{2a}{f(x)dx}=\int\limits_{0}^{a}{f(x)dx}+\int\limits_{0}^{a}{f(2a-x)dx} 0 ∫ 2 a f ( x ) d x = 0 ∫ a f ( x ) d x + 0 ∫ a f ( 2 a − x ) d x = 0 i f f ( 2 a − x ) = − f ( x ) =0\,\,if\,\,f(2a-x)=-f(x) = 0 i f f ( 2 a − x ) = − f ( x ) = 2 ∫ 0 a f ( x ) i f f ( 2 a − x ) = f ( x ) =2\int\limits_{0}^{a}{f(x)}\,\,if\,\,f(2a-x)=f(x) = 2 0 ∫ a f ( x ) i f f ( 2 a − x ) = f ( x ) ∫ − a a f ( x ) d x = { 0 i f f ( x ) i s o d d 2 ∫ 0 a f ( x ) d x i f f ( x ) i s o d d \int\limits_{-a}^{a}{f(x)dx=\left\{ \begin{matrix} 0 & if\,\,f\left( x \right)\,is\,odd \\ 2\int\limits_{0}^{a}{f\left( x \right)dx} & if\,\,f\left( x \right)\,is\,odd \\ \end{matrix} \right.\,\,\,\,\,\,} − a ∫ a f ( x ) d x = ⎩ ⎪ ⎨ ⎪ ⎧ 0 2 0 ∫ a f ( x ) d x i f f ( x ) i s o d d i f f ( x ) i s o d d
Problems on Integration Illustration:
∫ 0 2 x 2 [ x ] d x = ∫ 0 1 x 2 [ x ] d x + ∫ 1 2 x 2 [ x ] d x \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} 0 ∫ 2 x 2 [ x ] d x = 0 ∫ 1 x 2 [ x ] d x + 1 ∫ 2 x 2 [ x ] d x = ∫ 0 1 x 2 . 0 d x + ∫ 1 2 x 2 [ 1 ] d x =\int\limits_{0}^{1}{{{x}^{2}}.0\,dx}+\int\limits_{1}^{2}{{{x}^{2}}\left[ 1 \right]dx} = 0 ∫ 1 x 2 . 0 d x + 1 ∫ 2 x 2 [ 1 ] d x = 0 + x 3 3 ∣ 1 2 =0+\left. \frac{{{x}^{3}}}{3} \right|_{1}^{2} = 0 + 3 x 3 ∣ ∣ ∣ ∣ 1 2 = 8 − 1 3 = 7 3 =\frac{8-1}{3}\,=\frac{7}{3} = 3 8 − 1 = 3 7 Illustration:
∫ π / 6 π / 3 sin x sin x + cos x d x = ∫ cos x sin x + cos x d x \int\limits_{{\pi }/{6}\;}^{{\pi }/{3}\;}{\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx=}\int{\frac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx} π / 6 ∫ π / 3 s i n x + c o s x s i n x d x = ∫ s i n x + c o s x c o s x d x x → π 2 − x x\to \frac{\pi }{2}-x x → 2 π − x
2 I = ∫ π / 6 π / 3 sin x + cos x sin x + cos x = ∫ π / 6 π / 3 1 d x = π 6 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} 2 I = π / 6 ∫ π / 3 s i n x + c o s x s i n x + c o s x = π / 6 ∫ π / 3 1 d x = 6 π I = π 12 I=\frac{\pi }{12} I = 1 2 π Illustration:
I = ∫ sin 100 x cos 99 x I=\int{{{\sin }^{100}}x{{\cos }^{99}}x} I = ∫ sin 1 0 0 x cos 9 9 x Here f ( 2 π − x ) = f ( x ) f(2\pi -x)=f(x) f ( 2 π − x ) = f ( x )
Or I = 2 ∫ 0 π sin 100 x cos 99 x I=2\int\limits_{0}^{\pi }{{{\sin }^{100}}x{{\cos }^{99}}x} I = 2 0 ∫ π sin 1 0 0 x cos 9 9 x = 2 ∫ 0 π sin 100 ( π − x ) cos 99 ( π − x ) =2\int\limits_{0}^{\pi }{{{\sin }^{100}}\left( \pi -x \right){{\cos }^{99}}\left( \pi -x \right)} = 2 0 ∫ π sin 1 0 0 ( π − x ) cos 9 9 ( π − x )
I = -I
I = 0
Illustration:
∫ − 5 5 x 3 = 0 \int\limits_{-5}^{5}{{{x}^{3}}=0} − 5 ∫ 5 x 3 = 0 f ( x ) f(x) f ( x )
Area
A = ∫ a b f ( x ) d x = ∫ a b y d x A=\int\limits_{a}^{b}{f(x)dx=\int\limits_{a}^{b}{y\,dx}} A = a ∫ b f ( x ) d x = a ∫ b y d x
A = ∫ a b ( f ( x ) − g ( x ) ) d x A=\int\limits_{a}^{b}{\left( f(x)-g(x) \right)}\,dx A = a ∫ b ( f ( x ) − g ( x ) ) d x Illustration:
Find area between y 2 = 12 x {{y}^{2}}=12x y 2 = 1 2 x x 2 = 12 y {{x}^{2}}=12y x 2 = 1 2 y
A = ∫ 0 12 12 x − x 2 12 d x A=\int\limits_{0}^{12}{\sqrt{12x}-\frac{{{x}^{2}}}{12}dx} A = 0 ∫ 1 2 1 2 x − 1 2 x 2 d x = 12 x 3 / 2 3 / 2 − x 2 12 d x =\frac{12{{x}^{{3}/{2}\;}}}{{3}/{2}\;}-\frac{{{x}^{2}}}{12}dx = 3 / 2 1 2 x 3 / 2 − 1 2 x 2 d x = 8 x 3 / 2 ∣ 0 12 − x 3 36 ∣ 0 12 =\left. 8{{x}^{{3}/{2}\;}} \right|_{0}^{12}-\left. \frac{{{x}^{3}}}{36} \right|_{0}^{12} = 8 x 3 / 2 ∣ ∣ ∣ 0 1 2 − 3 6 x 3 ∣ ∣ ∣ ∣ 0 1 2 = 8 . 12 3 / 2 − 144 3 =8\,\,.\,\,{{12}^{{3}/{2}\;}}-\frac{144}{3} = 8 . 1 2 3 / 2 − 3 1 4 4 Linear Differential Equation Equation of form d y d x + p y = θ \frac{dy}{dx}+py=\theta d x d y + p y = θ
Ar solve by multiplying integrating function
If e ∫ p d x {{e}^{\int{pdx}}} e ∫ p d x
To get final solution of
y ( i f ) = ∫ Q ( I F ) + c y(if)=\int{Q(IF)+c} y ( i f ) = ∫ Q ( I F ) + c Illustration
Solve x 2 d y d x + y + 1 = 1 {{x}^{2}}\frac{dy}{dx+y+1}=1 x 2 d x + y + 1 d y = 1
Ans : Given equation is
d y d x + 1 x 2 y = 1 x x 2 \frac{dy}{dx}+\frac{1}{{{x}^{2}}}y=\frac{1x}{{{x}^{2}}} d x d y + x 2 1 y = x 2 1 x
p = 1 x 2 , q = 1 x 2 p=\frac{1}{{{x}^{2}}},q=\frac{1}{{{x}^{2}}} p = x 2 1 , q = x 2 1 IF= e ∫ 1 x x d x = e 1 x ={{e}^{\int{\frac{1}{{{x}^{x}}}dx}}}=e\frac{1}{x} = e ∫ x x 1 d x = e x 1
Ans:
y e − 1 x = ∫ p − 1 x 2 ( 1 x 2 ) d x + c y\,e\,\frac{-1}{x}=\int{{{p}^{\frac{-1}{{{x}^{2}}}}}}(\frac{1}{{{x}^{2}}})dx+c y e x − 1 = ∫ p x 2 − 1 ( x 2 1 ) d x + c y = 1 + c e 1 x y=1+c{{e}^{\frac{1}{x}}} y = 1 + c e x 1 General Form
We use combinations of different terms to get united term
Following terms are useful
1. x d y + y d x = d ( x y ) xdy+ydx=d(xy) x d y + y d x = d ( x y )
2. x d y − y d x x 2 = d ( y x ) \frac{xdy-ydx}{{{x}^{2}}}=d\left( \frac{y}{x} \right) x 2 x d y − y d x = d ( x y )
3. y d x − x d y y 2 = d ( x y ) \frac{ydx-xdy}{{{y}^{2}}}=d\left( \frac{x}{y} \right) y 2 y d x − x d y = d ( y x )
4. x d y + y d x x y = d ( l a x y ) \frac{xdy+ydx}{xy}=d(la\,xy) x y x d y + y d x = d ( l a x y )
5. x d y − y d x x y = d ( 1 ( y x ) ) \frac{xdy-ydx}{xy}=d\left( 1\left( \frac{y}{x} \right) \right) x y x d y − y d x = d ( 1 ( x y ) )
6. x d y − y d x x 2 + y 2 = d ( tan − 1 ( y x ) ) \frac{xdy-ydx}{{{x}^{2}}+{{y}^{2}}}=d\left( {{\tan }^{-1}}\left( \frac{y}{x} \right) \right) x 2 + y 2 x d y − y d x = d ( tan − 1 ( x y ) )
Illustration
Solve x d x + y d y = x d y − y d x x 2 + y 2 xdx+ydy=\frac{xdy-ydx}{{{x}^{2}}+{{y}^{2}}} x d x + y d y = x 2 + y 2 x d y − y d x
Ans:
1 2 d ( x 2 + y 2 ) = d ( tan − 1 ( y x ) ) \frac{1}{2}d({{x}^{2}}+{{y}^{2}})=d\left( {{\tan }^{-1}}\left( \frac{y}{x} \right) \right) 2 1 d ( x 2 + y 2 ) = d ( tan − 1 ( x y ) ) Integrating we get
1 2 ( x 2 + y 2 ) = tan − 1 ( y x ) + c \frac{1}{2}({{x}^{2}}+{{y}^{2}})={{\tan }^{-1}}\left( \frac{y}{x} \right)+c 2 1 ( x 2 + y 2 ) = tan − 1 ( x y ) + c Leibnitz’s rule f cont [a, b] & u ( x ) , v ( x ) u\left( x \right),v\left( x \right) u ( x ) , v ( x )
d d x ∫ u ( x ) v ( x ) f ( t ) d t = f ( v ( x ) ) d v ( x ) d x − f ( u ( x ) ) u ’ ( x ) \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) d x d u ( x ) ∫ v ( x ) f ( t ) d t = f ( v ( x ) ) d x d v ( x ) − f ( u ( x ) ) u ’ ( x ) Proof:
d d x F ( x ) = f ( x ) \frac{d}{dx}F\left( x \right)=f\left( x \right) d x d F ( x ) = f ( x ) ∴ ∫ u ( x ) v ( x ) f ( t ) d t = F ( v ( x ) ) − F ( u ( x ) ) \int\limits_{u\left( x \right)}^{v\left( x \right)}{f\left( t \right)dt=F\left( v\left( x \right) \right)}-F\left( u\left( x \right) \right) u ( x ) ∫ v ( x ) f ( t ) d t = F ( v ( x ) ) − F ( u ( x ) )
Practice Problems Q. ∫ x 2 x 3 1 log t d t = y \int\limits_{{{x}^{2}}}^{{{x}^{3}}}{\frac{1}{\log t}dt=y} x 2 ∫ x 3 l o g t 1 d t = y d y d x = x ( x − 1 ) ( log x ) − 1 \frac{dy}{dx}=x\left( x-1 \right){{\left( \log x \right)}^{-1}} d x d y = x ( x − 1 ) ( log x ) − 1
Q. If ∫ sin x 1 t 2 f ( t ) d t = 1 − sin x . \int\limits_{\sin x}^{1}{{{t}^{2}}f\left( t \right)dt=1-\sin x.} s i n x ∫ 1 t 2 f ( t ) d t = 1 − sin x . x ∈ ( 0 , π 2 ) x\in \left( 0,\frac{\pi }{2} \right) x ∈ ( 0 , 2 π ) f ( 1 3 ) = 3 f\left( \frac{1}{\sqrt{3}} \right)=3 f ( 3 1 ) = 3
Q. f ( 2 ) = 6 , f ’ ( 2 ) = 1 48 f\left( 2 \right)=6,f’\left( 2 \right)=\frac{1}{48} f ( 2 ) = 6 , f ’ ( 2 ) = 4 8 1 lim x → 2 ∫ 6 f ( x ) 4 t 3 x − 2 d t = 18 \underset{x\to 2}{\mathop{\lim }}\,\int\limits_{6}^{f\left( x \right)}{\frac{4{{t}^{3}}}{x-2}}dt=18 x → 2 lim 6 ∫ f ( x ) x − 2 4 t 3 d t = 1 8
Q. lim x → ∞ ∫ 0 x e x 2 d x ∫ 0 x e 2 x 2 d x = 0 \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 x → ∞ lim 0 ∫ x e 2 x 2 d x 0 ∫ x e x 2 d x = 0
Q. y = ∫ 1 8 sin 2 x sin − 1 t d t + ∫ 1 8 cos 2 x cos − 1 t d t . y=\int\limits_{\frac{1}{8}}^{{{\sin }^{2}}x}{{{\sin }^{-1}}\sqrt{t}\,dt}+\int\limits_{\frac{1}{8}}^{{{\cos }^{2}}x}{{{\cos }^{-1}}\sqrt{t}\,dt}. y = 8 1 ∫ s i n 2 x sin − 1 t d t + 8 1 ∫ c o s 2 x cos − 1 t d t . x ∈ [ 0 , π 2 ] x\in \left[ 0,\frac{\pi }{2} \right] x ∈ [ 0 , 2 π ]
∴ ( d y d x = 0 ) y = c o n s t \left( \frac{dy}{dx}=0 \right)y=const ( d x d y = 0 ) y = c o n s t x = π 4 y = 3 π 16 x=\frac{\pi }{4}y=\frac{3\pi }{16} x = 4 π y = 1 6 3 π
Q. f : ( 0 , ∞ ) → ( 0 , ∞ ) f:\left( 0,\infty \right)\to \left( 0,\infty \right) f : ( 0 , ∞ ) → ( 0 , ∞ ) x ∫ 0 x ( 1 − t ) f ( t ) d t = ∫ 0 x t f ( t ) d t x ∈ R + x\int\limits_{0}^{x}{\left( 1-t \right)f\left( t \right)dt=\int\limits_{0}^{x}{t\,f\left( t \right)dt}}\,\,\,\,\,\,x\in {{R}^{+}} x 0 ∫ x ( 1 − t ) f ( t ) d t = 0 ∫ x t f ( t ) d t x ∈ R +
Given f ( 1 ) = 1. f\left( 1 \right)=1. f ( 1 ) = 1 . f ( x ) . f ( x ) = 1 x 3 e ( 1 − 1 x ) f\left( x \right).\,\,\,\,\,\,\,f\left( x \right)=\frac{1}{{{x}^{3}}}{{e}^{\left( 1-\frac{1}{x} \right)}} f ( x ) . f ( x ) = x 3 1 e ( 1 − x 1 )
Hint differentiate twice. ∫ f ’ x f ( x ) = ∫ 1 − 3 x x 2 \int{\frac{f’x}{f\left( x \right)}}=\int{\frac{1-3x}{{{x}^{2}}}} ∫ f ( x ) f ’ x = ∫ x 2 1 − 3 x
Q. lim x → 4 ∫ 4 x 4 t − f ( t ) x − 4 d t = 16 − f ( 4 ) \underset{x\to 4}{\mathop{\lim }}\,\int\limits_{4}^{x}{\frac{4t-f\left( t \right)}{x-4}dt=16-f\left( 4 \right)} x → 4 lim 4 ∫ x x − 4 4 t − f ( t ) d t = 1 6 − f ( 4 )
Q. lim x → 0 ∫ 0 x cos t 2 d t x = 1 \underset{x\to 0}{\mathop{\lim }}\,\frac{\int\limits_{0}^{x}{\cos {{t}^{2}}dt}}{x}=1 x → 0 lim x 0 ∫ x c o s t 2 d t = 1
Q. Find points of minima for f ( x ) = ∫ 0 x t ( t − 1 ) ( t − 2 ) d t f\left( x \right)=\int\limits_{0}^{x}{t\left( t-1 \right)\left( t-2 \right)dt} f ( x ) = 0 ∫ x t ( t − 1 ) ( t − 2 ) d t n = 0 , 2 n=0,2 n = 0 , 2
Type
∫ a cos x + b sin x + p c cos x + d sin x + q d x , ∫ a e x + b e − x + c d e x + f e − x + h d x \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 ∫ c c o s x + d s i n x + q a c o s x + b s i n x + p d x , ∫ d e x + f e − x + h a e x + b e − x + c d x This is solved by N r = n D r + m D r ′ Nr=nDr+mDr' N r = n D r + m D r ′
Illustration:
∫ 2 cos x + 3 sin x 4 cos x + 5 sin x d x \int{\frac{2\cos x+3\sin x}{4\cos x+5\sin x}}\,dx ∫ 4 c o s x + 5 s i n x 2 c o s x + 3 s i n x d x Let 2 cos x + 3 sin x = a ( 4 cos x + 5 sin x ) + b ( − 4 sin x + 5 cos x ) 2\cos x+3\sin x=a\left( 4\cos x+5\sin x \right)+b\left( -4\sin x+5\cos x \right) 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
a = 23 41 b = − 2 41 a=\frac{23}{41}\,\,\,\,\,\,b=\frac{-2}{41} a = 4 1 2 3 b = 4 1 − 2 ∴ I = ∫ 23 41 − 2 41 ( − 4 sin x + 5 cos x 4 cos x + 5 sin x ) d x I=\int{\frac{23}{41}-\frac{2}{41}\left( \frac{-4\sin x+5\cos x}{4\cos x+5\sin x} \right)dx} I = ∫ 4 1 2 3 − 4 1 2 ( 4 c o s x + 5 s i n x − 4 s i n x + 5 c o s x ) d x = 23 41 x − 2 41 ℓ n ∣ 4 cos x + 5 sin x ∣ + c =\frac{23}{41}x-\frac{2}{41}\ell n\left| 4\cos x+5\sin x \right|+c = 4 1 2 3 x − 4 1 2 ℓ n ∣ 4 cos x + 5 sin x ∣ + c
By Parts Integration
∫ u v d x = u ∫ v d x − ∫ u ’ ( ∫ v d x ) d x \int{uv\,dx}=u\int{vdx}-\int{u’\left( \int{vdx} \right)}\,dx ∫ u v d x = u ∫ v d x − ∫ u ’ ( ∫ v d x ) d x Illustration:
Q. ∫ ℓ n x = ∫ ℓ n x . 1 d x \int{\ell n\,x}=\int{\ell n\,\,x.1\,\,dx} ∫ ℓ n x = ∫ ℓ n x . 1 d x = x ℓ n x − ∫ 1 x . x d x =x\,\ell n\,x-\int{\frac{1}{x}.\,x}\,dx = x ℓ n x − ∫ x 1 . x d x = x ℓ n x − x =x\,\ell n\,x-x = x ℓ n x − x
Q. ∫ x e x d x = x ∫ e x d x − ∫ ( x ) 1 ( ∫ e x d x ) d x \int{x\,{{e}^{x}}dx=x\int{{{e}^{x}}dx-\int{{{\left( x \right)}^{1}}\left( \int{{{e}^{x}}dx} \right)}dx}} ∫ x e x d x = x ∫ e x d x − ∫ ( x ) 1 ( ∫ e x d x ) d x = x e x − ∫ e x d x =x{{e}^{x}}-\int{{{e}^{x}}dx} = x e x − ∫ e x d x = x e x − e x =x{{e}^{x}}-{{e}^{x}} = x e x − e x
Q. ∫ ( x 3 + 3 x ) sin 3 x d x = ( x 3 + 3 x ) ( − cos 3 x 3 ) − ( 3 x 2 + 3 ) ( − sin 3 x 3 2 ) + 6 x ( cos 3 x 3 3 ) − 6 ( sin 3 x 3 4 ) + c \int{\left( {{x}^{3}}+3x \right)\sin 3x\,dx=\left( {{x}^{3}}+3x \right)\left( \frac{-\cos 3x}{3} \right)}-\left( 3{{x}^{2}}+3 \right)\left( \frac{-{{\sin }^{3}}x}{{{3}^{2}}} \right)+6x\left( \frac{\cos 3x}{{{3}^{3}}} \right)-6\left( \frac{{{\sin }^{3}}x}{{{3}^{4}}} \right)+c ∫ ( x 3 + 3 x ) sin 3 x d x = ( x 3 + 3 x ) ( 3 − c o s 3 x ) − ( 3 x 2 + 3 ) ( 3 2 − s i n 3 x ) + 6 x ( 3 3 c o s 3 x ) − 6 ( 3 4 s i n 3 x ) + c
Integration of Irrational Algebraic Functions Type ∫ d x ( a x + b ) k p x + q \int{\frac{dx}{{{\left( ax+b \right)}^{k}}\sqrt{px+q}}} ∫ ( a x + b ) k p x + q d x
Q. ∫ x ( x − 3 ) x + 1 d x \int{\frac{x}{\left( x-3 \right)\sqrt{x+1}}dx} ∫ ( x − 3 ) x + 1 x d x
Put x + 1 = t 2 , x+1={{t}^{2}}, x + 1 = t 2 ,
I = ∫ ( t 2 − 1 ) 2 t d t ( t 2 − 4 ) t = 2 ∫ t 2 − 1 t 2 − 4 d t 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} I = ∫ ( t 2 − 4 ) t ( t 2 − 1 ) 2 t d t = 2 ∫ t 2 − 4 t 2 − 1 d t = 2 ∫ 1 + 3 t 2 − 4 d t =2\int{1}+\frac{3}{{{t}^{2}}-4}dt = 2 ∫ 1 + t 2 − 4 3 d t = 2 t + 3 2 ℓ n ∣ t − 2 t + 2 ∣ + c =2t+\frac{3}{2}\ell n\left| \frac{t-2}{t+2} \right|+c = 2 t + 2 3 ℓ n ∣ ∣ ∣ t + 2 t − 2 ∣ ∣ ∣ + c = 2 x + 1 + 3 2 ℓ n ∣ x + 1 − 2 x + 1 + 2 ∣ + c =2\sqrt{x+1}+\frac{3}{2}\ell n\left| \frac{\sqrt{x+1}-2}{\sqrt{x+1}+2} \right|+c = 2 x + 1 + 2 3 ℓ n ∣ ∣ ∣ ∣ x + 1 + 2 x + 1 − 2 ∣ ∣ ∣ ∣ + c ⇒ ∫ 0 2 a f ( x ) d x = ∫ 0 a f ( x ) d x + ∫ 0 a f ( 2 a − x ) d x \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} 0 ∫ 2 a f ( x ) d x = 0 ∫ a f ( x ) d x + 0 ∫ a f ( 2 a − x ) d x
= 0 if f ( 2 a − x ) = − f ( x ) f\left( 2a-x \right)=-f\left( x \right) f ( 2 a − x ) = − f ( x ) = 2 ∫ 0 a f ( x ) =2\int\limits_{0}^{a}{f\left( x \right)} = 2 0 ∫ a f ( x ) f ( 2 a − x ) = f ( x ) f\left( 2a-x \right)=f\left( x \right) f ( 2 a − x ) = f ( x )
Illustration:
Q. I = ∫ sin 100 x cos 99 x I=\int{{{\sin }^{100}}x{{\cos }^{99}}x} I = ∫ sin 1 0 0 x cos 9 9 x
Here f ( 2 π − x ) = f ( x ) f\left( 2\pi -x \right)=f\left( x \right) f ( 2 π − x ) = f ( x )
Or I = 2 ∫ 0 π sin 100 x cos 99 x I=2\int\limits_{0}^{\pi }{{{\sin }^{100}}x{{\cos }^{99}}x} I = 2 0 ∫ π sin 1 0 0 x cos 9 9 x = 2 ∫ 0 π sin 100 ( π − x ) cos 99 ( π − x ) =2\int\limits_{0}^{\pi }{{{\sin }^{100}}\left( \pi -x \right)}{{\cos }^{99}}\left( \pi -x \right) = 2 0 ∫ π sin 1 0 0 ( π − x ) cos 9 9 ( π − x )
I = -I
I = 0
⇒ ∫ − a a f ( x ) d x = { 0 i f f ( x ) i s o d d 2 ∫ 0 a f ( x ) d x i f f ( x ) i s e v e n \int\limits_{-a}^{a}{f\left( x \right)dx=\left\{ \begin{matrix} 0 & if\,\,f\left( x \right)\,\,is\,\,odd \\ 2\int\limits_{0}^{a}{f\left( x \right)dx} & if\,\,f\left( x \right)\,\,is\,\,even \\ \end{matrix} \right.} − a ∫ a f ( x ) d x = ⎩ ⎪ ⎨ ⎪ ⎧ 0 2 0 ∫ a f ( x ) d x i f f ( x ) i s o d d i f f ( x ) i s e v e n
Illustration:
Q. ∫ − 5 5 x 3 = 0 \int\limits_{-5}^{5}{{{x}^{3}}=0} − 5 ∫ 5 x 3 = 0 f ( x ) f\left( x \right) f ( x )
Q. Find area between y 2 ≤ 4 x , x 2 + y 2 ≥ 2 x {{y}^{2}}\le 4x,{{x}^{2}}+{{y}^{2}}\ge 2x y 2 ≤ 4 x , x 2 + y 2 ≥ 2 x x ≤ y + 2. x\le y+2. x ≤ y + 2 .
Answer:
A = ∫ 0 ( 3 + 1 ) 2 4 x d x − A=\int\limits_{0}^{{{\left( \sqrt{3}+1 \right)}^{2}}}{\sqrt{4x}\,\,dx-} A = 0 ∫ ( 3 + 1 ) 2 4 x d x −
= 4 [ 2 x 3 / 2 3 ] 0 ( 3 + 1 ) 2 − π 2 − 1 2 ( 3 + 1 ) 2 2 ( 3 + 1 ) =\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) = 4 [ 3 2 x 3 / 2 ] 0 ( 3 + 1 ) 2 − 2 π − 2 1 ( 3 + 1 ) 2 2 ( 3 + 1 ) = ( 3 + 1 ) 3 3 − π 2 =\frac{{{\left( \sqrt{3}+1 \right)}^{3}}}{3}-\frac{\pi }{2} = 3 ( 3 + 1 ) 3 − 2 π Optimisation of area for greatest and least values Illustration:
If area by y = f ( x ) y=f\left( x \right) y = f ( x ) y = x 2 + 2 y={{x}^{2}}+2 y = x 2 + 2 x = 2 & x = α x=2\And x=\alpha x = 2 & x = α α 3 − 4 α 2 + 8. {{\alpha }^{3}}-4{{\alpha }^{2}}+8. α 3 − 4 α 2 + 8 . f ( x ) . f\left( x \right). f ( x ) .
Answer:
α 3 − 4 α 2 + 8 = ∫ 2 α ( x 2 + 2 − f ( x ) ) d x {{\alpha }^{3}}-4{{\alpha }^{2}}+8=\int\limits_{2}^{\alpha }{\left( {{x}^{2}}+2-f\left( x \right) \right)dx} α 3 − 4 α 2 + 8 = 2 ∫ α ( x 2 + 2 − f ( x ) ) d x Differentiating with Labniz equation
3 α 2 − 8 α = α 2 + 2 − f ( α ) 3{{\alpha }^{2}}-8\alpha ={{\alpha }^{2}}+2-f\left( \alpha \right) 3 α 2 − 8 α = α 2 + 2 − f ( α ) f ( x ) = − 2 x 2 + 8 x + 2 f\left( x \right)=-2{{x}^{2}}+8x+2 f ( x ) = − 2 x 2 + 8 x + 2
