Indefinite Integration Integration means summation. Integration originated during the course of finding the area of a plane figure. Integration is the reverse of differentiation. It is also called as antiderivative. In this section, aspirants will learn the list of important formulas, how to use integral properties to solve integration problems, integration methods and many more.
The process of finding an indefinite integral of a given function is called integration of the function. Integrating a function f(x) means finding a function Φ ( x ) s u c h t h a t d d x ( Φ ( x ) ) = f ( x ) ) \Phi (x) \: such \: that \frac{d}{dx}\left ( \Phi (x) \right ) = f(x)) Φ ( x ) s u c h t h a t d x d ( Φ ( x ) ) = f ( x ) )
Standard formulas for Indefinite Integration
∫ x n d x = x n + 1 n + 1 + c \int{{{x}^{n}}dx=\frac{{{x}^{n+1}}}{n+1}}+c ∫ x n d x = n + 1 x n + 1 + c ∫ a x d x = a x ℓ n a + c \int{{{a}^{x}}dx}=\frac{{{a}^{x}}}{\ell n\,a}+c ∫ a x d x = ℓ n a a x + c ∫ e x d x = e x + c \int{{{e}^{x}}}dx={{e}^{x}}+c ∫ e x d x = e x + c ∫ 1 x d x = ℓ n x + c \int{\frac{1}{x}}dx=\ell n\,x+c ∫ x 1 d x = ℓ n x + c ∫ sin x d x = − cos x + c \int{\sin x\,dx=-\cos x+c} ∫ sin x d x = − cos x + c ∫ cos x d x = sin x + c \int{\cos x\,dx=\sin x+c} ∫ cos x d x = sin x + c ∫ sec 2 x d x = tan x + c \int{{{\sec }^{2}}x\,\,dx=\tan x+c} ∫ sec 2 x d x = tan x + c ∫ cos e c 2 x d x = − cot x + c \int{\cos e{{c}^{2}}x\,\,dx=-\cot x\,+c} ∫ cos e c 2 x d x = − cot x + c ∫ sec x tan x d x = sec x + c \int{\sec x\tan xdx=\sec x+c} ∫ sec x tan x d x = sec x + c ∫ cos e c x cot x d x = − cos e c x + c \int{\cos ec\,\,x\,\,}\cot x\,dx=-\cos ec\,x+c ∫ cos e c x cot x d x = − cos e c x + c ∫ 1 1 − x 2 d x = = { sin − 1 x + c − cos − 1 x + c \int{\frac{1}{\sqrt{1-{{x}^{2}}}}dx\begin{matrix} = \\ = \\ \end{matrix}}\left\{ \begin{matrix} {{\sin }^{-1}}x+c \\ -{{\cos }^{-1}}x+c \\ \end{matrix} \right. ∫ 1 − x 2 1 d x = = { sin − 1 x + c − cos − 1 x + c ∫ 1 1 + x 2 d x = = { tan − 1 x + c − cot − 1 x + c \int{\frac{1}{1+{{x}^{2}}}}dx\begin{matrix} = \\ = \\ \end{matrix}\left\{ \begin{matrix} {{\tan }^{-1}}x+c \\ -{{\cot }^{-1}}x+c \\ \end{matrix} \right. ∫ 1 + x 2 1 d x = = { tan − 1 x + c − cot − 1 x + c 1 x x 2 − 1 = { sec − 1 x + c − cos e c − 1 x + c \frac{1}{x\sqrt{{{x}^{2}}-1}}=\left\{ \begin{matrix} {{\sec }^{-1}}x+c \\ -\cos e{{c}^{-1}}x+c \\ \end{matrix} \right. x x 2 − 1 1 = { sec − 1 x + c − cos e c − 1 x + c
⇒ Also Read – Introduction to Integration
Properties of Indefinite Integration
k ∫ f ( x ) d x = k ∫ f ( x ) k\int{f\left( x \right)dx=k\int{f\left( x \right)}} k ∫ f ( x ) d x = k ∫ f ( x ) ∫ ( f ( x ) + g ( x ) ) d x = ∫ f ( x ) d x + ∫ g ( x ) d x \int{\left( f\left( x \right)+g\left( x \right) \right)dx=\int{f\left( x \right)dx+\int{g\left( x \right)dx}}} ∫ ( f ( x ) + g ( x ) ) d x = ∫ f ( x ) d x + ∫ g ( x ) d x
Below are some Illustrations based on integration properties:
Illustration 1: Solve ∫ ( 1 + x ) 3 x 2 3 d x \int{\frac{{{\left( 1+x \right)}^{3}}}{\sqrt[3]{{{x}^{2}}}}}dx ∫ 3 x 2 ( 1 + x ) 3 d x
Solution: ∫ ( 1 + x ) 3 x 2 3 d x = ∫ 1 + 3 x + 3 x 2 + x 3 x 2 / 3 d x \int{\frac{{{\left( 1+x \right)}^{3}}}{\sqrt[3]{{{x}^{2}}}}}dx=\int{\frac{1+3x+3{{x}^{2}}+{{x}^{3}}}{{{x}^{2/3}}}}dx ∫ 3 x 2 ( 1 + x ) 3 d x = ∫ x 2 / 3 1 + 3 x + 3 x 2 + x 3 d x = ∫ x 2 / 3 + 3 x 1 / 3 + 3 x 4 / 3 + x 7 / 3 d x =\int{{{x}^{2/3}}+3{{x}^{1/3}}+3{{x}^{4/3}}}+{{x}^{7/3}}dx = ∫ x 2 / 3 + 3 x 1 / 3 + 3 x 4 / 3 + x 7 / 3 d x = 3 x 1 / 3 + 9 4 x 4 / 3 + 9 7 x 7 / 3 + 3 10 x 10 / 3 + c =3{{x}^{1/3}}+\frac{9}{4}{{x}^{4/3}}+\frac{9}{7}{{x}^{7/3}}+\frac{3}{10}{{x}^{10/3}}+c = 3 x 1 / 3 + 4 9 x 4 / 3 + 7 9 x 7 / 3 + 1 0 3 x 1 0 / 3 + c
Illustration 2: Solve ∫ sec 2 x cos e c 2 x d x \int{{\sec }^{2}}x\cos e{{c}^{2}}x\,\,dx ∫ sec 2 x cos e c 2 x d x
Solution: ∫ sec 2 x cos e c 2 x d x = ∫ sin 2 x + cos 2 x sin 2 x cos 2 x d x \int{{{\sec }^{2}}x\cos e{{c}^{2}}x\,\,dx=\int{\frac{{{\sin }^{2}}x+{{\cos }^{2}}x}{{{\sin }^{2}}x\,{{\cos }^{2}}x}}dx} ∫ sec 2 x cos e c 2 x d x = ∫ s i n 2 x c o s 2 x s i n 2 x + c o s 2 x d x = ∫ sec 2 x + cos e c 2 x d x =\int{{{\sec }^{2}}x+\cos e{{c}^{2}}x\,\,dx} = ∫ sec 2 x + cos e c 2 x d x = tan x − cot x + c =\tan x-\cot x+c = tan x − cot x + c
Standard integral form f ( a x + b ) f\left( ax+b \right) f ( a x + b ) ∫ ( a x + b ) n d x = ( a x + b ) n + 1 ( n + 1 ) a \int{{{\left( ax+b \right)}^{n}}dx=\frac{{{\left( ax+b \right)}^{n+1}}}{\left( n+1 \right)a}} ∫ ( a x + b ) n d x = ( n + 1 ) a ( a x + b ) n + 1 ∫ 1 a x + b = d x = ℓ n ∣ a x + b ∣ a + c \int{\frac{1}{ax+b}}=dx=\frac{\ell n\left| ax+b \right|}{a}+c ∫ a x + b 1 = d x = a ℓ n ∣ a x + b ∣ + c
and so on…
we follow as below
If ∫ f ( x ) = F ( x ) + c \int{f\left( x \right)}=F\left( x \right)+c ∫ f ( x ) = F ( x ) + c ∫ f ( a x + b ) = F ( a x + b ) a + c \int{f\left( ax+b \right)}=\frac{F\left( ax+b \right)}{a}+c ∫ f ( a x + b ) = a F ( a x + b ) + c
Methods of Integration
Substitution
Integration by partial fractions
By parts
Euler substitution
Reduction method
Substitution method A new variable is introduced such that direct integration formula can be applied.
Problems on Indefinite Integration Problem 1. ∫ 2 x sin ( x 2 ) d x \int{2x\,\,\sin \left( {{x}^{2}} \right)dx} ∫ 2 x sin ( x 2 ) d x
Solution: Put x 2 = t {{x}^{2}}=t x 2 = t
the 2 x d x = d t 2x\,\,dx=dt 2 x d x = d t ∫ sin t d t = − cos t + c \int{\sin t\,\,dt=-\cos t+c} ∫ sin t d t = − cos t + c = − cos x 2 + c =-\cos {{x}^{2}}+c = − cos x 2 + c
Problem 2. ∫ sin 3 x cos 5 x d x \int{{{\sin }^{3}}x{{\cos }^{5}}x\,\,dx} ∫ sin 3 x cos 5 x d x
Solution: = ∫ sin 2 x cos 5 x sin x d x =\int{{{\sin }^{2}}x{{\cos }^{5}}x\sin x\,\,dx} = ∫ sin 2 x cos 5 x sin x d x = ∫ ( 1 − cos 2 x ) cos 5 x sin x d x =\int{\left( 1-{{\cos }^{2}}x \right)}{{\cos }^{5}}x\,\,\sin x\,\,dx = ∫ ( 1 − cos 2 x ) cos 5 x sin x d x
put cos x = t \cos x=t cos x = t
∴ − sin x d x = d t \,\,\,-\sin x\,dx=dt − sin x d x = d t = ∫ ( 1 − t 2 ) t 5 ( − d t ) =\int{\left( 1-{{t}^{2}} \right)}{{t}^{5}}\left( -dt \right) = ∫ ( 1 − t 2 ) t 5 ( − d t ) = ∫ t 5 − t 7 d t =\int{{{t}^{5}}-{{t}^{7}}}dt = ∫ t 5 − t 7 d t = − ( t 6 6 − t 8 8 ) =-\left( \frac{{{t}^{6}}}{6}-\frac{{{t}^{8}}}{8} \right) = − ( 6 t 6 − 8 t 8 ) = − cos 6 x 6 + cos 8 x 8 + c =-\frac{{{\cos }^{6}}x}{6}+\frac{{{\cos }^{8}}x}{8}+c = − 6 c o s 6 x + 8 c o s 8 x + c
Note: Type I
∫ sin m x cos n x d x \int{{{\sin }^{m}}x{{\cos }^{n}}x\,\,dx} ∫ sin m x cos n x d x Rule:
If m, n one of them odd then substitute for even power.
If both odd then substitute either of them.
If both even use trigonometric identities.
Type II: Substitution of Trigonometric functions.
a 2 − x 2 → x = a sin θ \sqrt{{{a}^{2}}-{{x}^{2}}}\rightarrow{{}}x=a\sin \theta a 2 − x 2 → x = a sin θ x 2 + a 2 → x = a tan θ \sqrt{{{x}^{2}}+{{a}^{2}}}\rightarrow{{}}x=a\tan \theta x 2 + a 2 → x = a tan θ x 2 − a 2 → x = a sec θ \sqrt{{{x}^{2}}-{{a}^{2}}}\rightarrow{{}}x=a\sec \theta x 2 − a 2 → x = a sec θ x − a b − x → x = a cos 2 θ + b sin 2 θ \sqrt{\frac{x-a}{b-x}}\rightarrow{{}}x=a{{\cos }^{2}}\theta +b{{\sin }^{2}}\theta b − x x − a → x = a cos 2 θ + b sin 2 θ x − a x − b → x = a sec 2 θ − b \sqrt{\frac{x-a}{x-b}}\rightarrow{{}}x=a{{\sec }^{2}}\theta -b x − b x − a → x = a sec 2 θ − b
Important formulae set for Indefinite Integration
∫ d x a 2 − x 2 = sin − 1 ( x a ) + c \int{\frac{dx}{\sqrt{{{a}^{2}}-{{x}^{2}}}}}={{\sin }^{-1}}\left( \frac{x}{a} \right)+c ∫ a 2 − x 2 d x = sin − 1 ( a x ) + c ∫ d x x 2 − a 2 = ℓ n ∣ x + x 2 − a 2 ∣ \int{\frac{dx}{\sqrt{{{x}^{2}}-{{a}^{2}}}}=\ell n\left| x+\sqrt{{{x}^{2}}-{{a}^{2}}} \right|} ∫ x 2 − a 2 d x = ℓ n ∣ ∣ ∣ x + x 2 − a 2 ∣ ∣ ∣ ∫ d x x 2 + a 2 = ℓ n ∣ x + x 2 + a 2 ∣ \int{\frac{dx}{\sqrt{{{x}^{2}}+{{a}^{2}}}}=\ell n\left| x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right|} ∫ x 2 + a 2 d x = ℓ n ∣ ∣ ∣ x + x 2 + a 2 ∣ ∣ ∣ ∫ 1 a 2 + x 2 d x = 1 a tan − 1 ( x a ) . \int{\frac{1}{{{a}^{2}}+{{x}^{2}}}}dx=\frac{1}{a}{{\tan }^{-1}}\left( \frac{x}{a} \right). ∫ a 2 + x 2 1 d x = a 1 tan − 1 ( a x ) . ∫ 1 a 2 − x 2 d x = 1 2 a ℓ n ∣ a + x a − x ∣ . \int{\frac{1}{{{a}^{2}}-{{x}^{2}}}}dx=\frac{1}{2a}\ell n\left| \frac{a+x}{a-x} \right|. ∫ a 2 − x 2 1 d x = 2 a 1 ℓ n ∣ ∣ ∣ a − x a + x ∣ ∣ ∣ . ∫ 1 x 2 − a 2 d x = 1 2 a ℓ n ∣ x − a x + a ∣ . \int{\frac{1}{{{x}^{2}}-{{a}^{2}}}}dx=\frac{1}{2a}\ell n\left| \frac{x-a}{x+a} \right|. ∫ x 2 − a 2 1 d x = 2 a 1 ℓ n ∣ ∣ ∣ x + a x − a ∣ ∣ ∣ . ∫ a 2 − x 2 d x = x 2 a 2 − x 2 + a 2 2 sin − 1 ( x a ) . \int{\sqrt{{{a}^{2}}-{{x}^{2}}}}dx=\frac{x}{2}\sqrt{{{a}^{2}}-{{x}^{2}}}+\frac{{{a}^{2}}}{2}{{\sin }^{-1}}\left( \frac{x}{a} \right). ∫ a 2 − x 2 d x = 2 x a 2 − x 2 + 2 a 2 sin − 1 ( a x ) . ∫ a 2 + x 2 d x = x 2 x 2 + a 2 + a 2 2 ℓ n ∣ x + x 2 + a 2 ∣ \int{\sqrt{{{a}^{2}}+{{x}^{2}}}}dx=\frac{x}{2}\sqrt{{{x}^{2}}+{{a}^{2}}}+\frac{{{a}^{2}}}{2}\ell n\left| x+\sqrt{{{x}^{2}}+{{a}^{2}}} \right| ∫ a 2 + x 2 d x = 2 x x 2 + a 2 + 2 a 2 ℓ n ∣ ∣ ∣ x + x 2 + a 2 ∣ ∣ ∣ ∫ x 2 − a 2 d x = x 2 x 2 − a 2 + a 2 2 ℓ n ∣ x + x 2 − a 2 ∣ \int{\sqrt{{{x}^{2}}-{{a}^{2}}}}dx=\frac{x}{2}\sqrt{{{x}^{2}}-{{a}^{2}}}+\frac{{{a}^{2}}}{2}\ell n\left| x+\sqrt{{{x}^{2}}-{{a}^{2}}} \right| ∫ x 2 − a 2 d x = 2 x x 2 − a 2 + 2 a 2 ℓ n ∣ ∣ ∣ x + x 2 − a 2 ∣ ∣ ∣
Problem Type 1:
∫ d x x 2 + 2 x + 2 = ∫ d x ( x + 1 ) 2 + 1 \int{\frac{dx}{\sqrt{{{x}^{2}}+2x+2}}}=\int{\frac{dx}{\sqrt{{{\left( x+1 \right)}^{2}}+1}}} ∫ x 2 + 2 x + 2 d x = ∫ ( x + 1 ) 2 + 1 d x = log ∣ ( x + 1 ) + ( x + 1 ) 2 + 1 ∣ . =\log \left| \left( x+1 \right)+\sqrt{{{\left( x+1 \right)}^{2}}+1} \right|. = log ∣ ∣ ∣ ∣ ∣ ( x + 1 ) + ( x + 1 ) 2 + 1 ∣ ∣ ∣ ∣ ∣ . Problem Type 2:
∫ 5 x 2 + 2 x + 3 d x \int{\sqrt{5{{x}^{2}}+2x+3}}\,\,dx ∫ 5 x 2 + 2 x + 3 d x here
5 x 2 + 2 x + 3 = 5 [ ( x + 1 5 ) 2 + ( 14 25 ) 2 ] 5{{x}^{2}}+2x+3=5\left[ {{\left( x+\frac{1}{5} \right)}^{2}}+{{\left( \sqrt{\frac{14}{25}} \right)}^{2}} \right] 5 x 2 + 2 x + 3 = 5 [ ( x + 5 1 ) 2 + ( 2 5 1 4 ) 2 ] ∫ 5 x 2 + 2 x + 3 d x = 5 ∫ ( x + 1 5 ) 2 + ( 14 25 ) 2 \int{\sqrt{5{{x}^{2}}+2x+3}\,dx}=\sqrt{5}\int{\sqrt{{{\left( x+\frac{1}{5} \right)}^{2}}+{{\left( \sqrt{\frac{14}{25}} \right)}^{2}}}} ∫ 5 x 2 + 2 x + 3 d x = 5 ∫ ( x + 5 1 ) 2 + ( 2 5 1 4 ) 2 = 5 ( x + 1 5 ) 2 ( x + 1 5 ) 2 + ( 14 15 ) 2 + 14 25.2 log ∣ ( x + 1 5 ) + ( x + 1 5 ) 2 + 14 15 ∣ =\sqrt{5}\frac{\left( x+\frac{1}{5} \right)}{2}\sqrt{{{\left( x+\frac{1}{5} \right)}^{2}}+{{\left( \sqrt{\frac{14}{15}} \right)}^{2}}}+\frac{14}{25.2}\log \left| \left( x+\frac{1}{5} \right)+\sqrt{{{\left( x+\frac{1}{5} \right)}^{2}}+\frac{14}{15}} \right| = 5 2 ( x + 5 1 ) ( x + 5 1 ) 2 + ( 1 5 1 4 ) 2 + 2 5 . 2 1 4 log ∣ ∣ ∣ ∣ ∣ ( x + 5 1 ) + ( x + 5 1 ) 2 + 1 5 1 4 ∣ ∣ ∣ ∣ ∣ Important forms
∫ p x + q a x 2 + b x + c d x & ∫ p x + q a x 2 + b x + c d x & ∫ ( p x + q ) a x 2 + b x + c d x \int{\frac{px+q}{a{{x}^{2}}+bx+c}}dx\And \int{\frac{px+q}{\sqrt{a{{x}^{2}}+bx+c}}}\,dx\And \int{\left( px+q \right)}\sqrt{a{{x}^{2}}+bx+c}\,dx ∫ a x 2 + b x + c p x + q d x & ∫ a x 2 + b x + c p x + q d x & ∫ ( p x + q ) a x 2 + b x + c d x We express p x + q px+q p x + q m ( a x 2 + b x + c ) 1 + n m{{\left( a{{x}^{2}}+bx+c \right)}^{1}}+n m ( a x 2 + b x + c ) 1 + n
Problem Type 3:
I = ∫ x + 1 x 2 + 3 x + 4 d x I=\int{\frac{x+1}{{{x}^{2}}+3x+4}}dx I = ∫ x 2 + 3 x + 4 x + 1 d x x + 1 = a ( x 2 + 3 x + 4 ) 1 + b x+1=a{{\left( {{x}^{2}}+3x+4 \right)}^{1}}+b x + 1 = a ( x 2 + 3 x + 4 ) 1 + b = 2 a x + ( 3 a + b ) =2ax+\left( 3a+b \right) = 2 a x + ( 3 a + b ) ∴ a = 1 2 b = − 1 2 a=\frac{1}{2}\,\,\,\,b=\frac{-1}{2} a = 2 1 b = 2 − 1
⇒ I = ∫ 1 2 ( x 2 + 3 x + 4 ) 1 x 2 + 3 x + 4 − 1 2 ( 1 x 2 + 3 x + 4 ) I=\int{\frac{1}{2}}\frac{{{\left( {{x}^{2}}+3x+4 \right)}^{1}}}{{{x}^{2}}+3x+4}-\frac{1}{2}\left( \frac{1}{{{x}^{2}}+3x+4} \right) I = ∫ 2 1 x 2 + 3 x + 4 ( x 2 + 3 x + 4 ) 1 − 2 1 ( x 2 + 3 x + 4 1 ) = 1 2 log ( x 2 + 3 x + 4 ) − 1 2 ∫ 1 ( x + 3 2 ) 2 + 7 4 =\frac{1}{2}\log \left( {{x}^{2}}+3x+4 \right)-\frac{1}{2}\int{\frac{1}{{{\left( x+\frac{3}{2} \right)}^{2}}+\frac{7}{4}}} = 2 1 log ( x 2 + 3 x + 4 ) − 2 1 ∫ ( x + 2 3 ) 2 + 4 7 1 = 1 2 log ( x 2 + 3 x + 4 ) − 1 2 1 7 4 tan − 1 ( x + 3 2 ) 7 4 =\frac{1}{2}\log \left( {{x}^{2}}+3x+4 \right)-\frac{1}{2}\frac{1}{\sqrt{\frac{7}{4}}}{{\tan }^{-1}}\frac{\left( x+\frac{3}{2} \right)}{\sqrt{\frac{7}{4}}} = 2 1 log ( x 2 + 3 x + 4 ) − 2 1 4 7 1 tan − 1 4 7 ( x + 2 3 ) = 1 2 log ∣ x 2 + 3 x + 4 ∣ − 1 7 tan − 1 ( 2 x + 3 7 ) =\frac{1}{2}\log \left| {{x}^{2}}+3x+4 \right|-\frac{1}{\sqrt{7}}{{\tan }^{-1}}\left( \frac{2x+3}{\sqrt{7}} \right) = 2 1 log ∣ ∣ ∣ x 2 + 3 x + 4 ∣ ∣ ∣ − 7 1 tan − 1 ( 7 2 x + 3 )
Biquadratic Substitutions 1. I = ∫ f ( x + 1 x ) ( 1 − 1 x 2 ) d x I=\int{f\left( x+\frac{1}{x} \right)}\left( 1-\frac{1}{{{x}^{2}}} \right)dx I = ∫ f ( x + x 1 ) ( 1 − x 2 1 ) d x
Put x + 1 x = t x+\frac{1}{x}=t x + x 1 = t
2. I = ∫ f ( x − 1 x ) ( 1 + 1 x 2 ) d x I=\int{f\left( x-\frac{1}{x} \right)}\left( 1+\frac{1}{{{x}^{2}}} \right)dx I = ∫ f ( x − x 1 ) ( 1 + x 2 1 ) d x
Put ( x − 1 x ) = t \left( x-\frac{1}{x} \right)=t ( x − x 1 ) = t
Problem Type 4:
∫ 1 + x 2 1 + x 4 d x = ∫ ( 1 + 1 x 2 ) ( 1 x 2 + x 2 ) d x \int{\frac{1+{{x}^{2}}}{1+{{x}^{4}}}}\,dx=\int{\frac{\left( 1+\frac{1}{{{x}^{2}}} \right)}{\left( \frac{1}{{{x}^{2}}}+{{x}^{2}} \right)}}\,dx ∫ 1 + x 4 1 + x 2 d x = ∫ ( x 2 1 + x 2 ) ( 1 + x 2 1 ) d x = ∫ ( 1 + 1 x 2 ) d x ( x − 1 x ) 2 + 2 =\int{\frac{\left( 1+\frac{1}{{{x}^{2}}} \right)dx}{{{\left( x-\frac{1}{x} \right)}^{2}}+2}} = ∫ ( x − x 1 ) 2 + 2 ( 1 + x 2 1 ) d x = ∫ d ( x − 1 x ) ( x − 1 x ) 2 + 2 =\int{\frac{d\left( x-\frac{1}{x} \right)}{{{\left( x-\frac{1}{x} \right)}^{2}}+2}} = ∫ ( x − x 1 ) 2 + 2 d ( x − x 1 ) = 1 2 tan − 1 ( ( x − 1 x ) 2 ) =\frac{1}{\sqrt{2}}{{\tan }^{-1}}\left( \frac{\left( x-\frac{1}{x} \right)}{\sqrt{2}} \right) = 2 1 tan − 1 ( 2 ( x − x 1 ) ) Integration by Partial Fraction Integrals of rational functions can be evaluated by splitting them into partial fractions.
f ( x ) g ( x ) \frac{f\left( x \right)}{g\left( x \right)} g ( x ) f ( x ) f & g f\And g f & g
If the degree of f < f< f <
Otherwise, it is an improper fraction.
Then f ( x ) g ( x ) = h ( x ) + d ( x ) g ( x ) \frac{f\left( x \right)}{g\left( x \right)}=h\left( x \right)+\frac{d\left( x \right)}{g\left( x \right)} g ( x ) f ( x ) = h ( x ) + g ( x ) d ( x )
Cases:
1. When g ( x ) g\left( x \right) g ( x )
g ( x ) = ( x − a 1 ) ( x − a 2 ) … . ( x − a n ) g\left( x \right)=\left( x-{{a}_{1}} \right)\left( x-{{a}_{2}} \right)….\left( x-{{a}_{n}} \right) g ( x ) = ( x − a 1 ) ( x − a 2 ) … . ( x − a n )
f g = A 1 x − a 1 + A 2 x − a 2 + … . A n x − a n . \frac{f}{g}=\frac{{{A}_{1}}}{x-{{a}_{1}}}+\frac{{{A}_{2}}}{x-{{a}_{2}}}+….\frac{{{A}_{n}}}{x-{{a}_{n}}}. g f = x − a 1 A 1 + x − a 2 A 2 + … . x − a n A n . 2. Some factors of g are repeating then g ( x ) = ( x − a ) k ( x − a 1 ) ( x − a 2 ) … g\left( x \right)={{\left( x-a \right)}^{k}}\left( x-{{a}_{1}} \right)\left( x-{{a}_{2}} \right)… g ( x ) = ( x − a ) k ( x − a 1 ) ( x − a 2 ) … f g = A 1 ( x − a ) + A 2 ( x − a ) 2 + … . . A k ( x − a ) n + … \frac{f}{g}=\frac{{{A}_{1}}}{\left( x-a \right)}+\frac{{{A}_{2}}}{{{\left( x-a \right)}^{2}}}+…..\frac{{{A}_{k}}}{{{\left( x-a \right)}^{n}}}+… g f = ( x − a ) A 1 + ( x − a ) 2 A 2 + … . . ( x − a ) n A k + …
3. g ( x ) g\left( x \right) g ( x ) f g = A x + B a x 2 + b x + c \frac{f}{g}=\frac{Ax+B}{a{{x}^{2}}+bx+c} g f = a x 2 + b x + c A x + B
Where A & B are constants determined by comparing coefficients.
∫ d x ( x + 1 ) ( x − 2 ) = ∫ a x + 1 + b x − 2 d x \int{\frac{dx}{\left( x+1 \right)\left( x-2 \right)}}=\int{\frac{a}{x+1}+\frac{b}{x-2}}\,dx ∫ ( x + 1 ) ( x − 2 ) d x = ∫ x + 1 a + x − 2 b d x = ∫ − 1 3 x + 1 + ∫ 1 2 x − 2 d x =\int{\frac{-\frac{1}{3}}{x+1}}+\int{\frac{\frac{1}{2}}{x-2}}\,dx = ∫ x + 1 − 3 1 + ∫ x − 2 2 1 d x = − 1 3 ℓ n ∣ x + 1 ∣ + 1 2 ℓ n ∣ x − 2 ∣ =-\frac{1}{3}\ell n\left| x+1 \right|+\frac{1}{2}\ell n\left| x-2 \right| = − 3 1 ℓ n ∣ x + 1 ∣ + 2 1 ℓ n ∣ x − 2 ∣ ∫ 2 x 2 − 1 ( x − 1 ) ( x + 1 ) 2 d x = ∫ A x − 1 + B ( x + 1 ) + C ( x + 1 ) 2 \int{\frac{2{{x}^{2}}-1}{\left( x-1 \right){{\left( x+1 \right)}^{2}}}dx=\int{\frac{A}{x-1}+\frac{B}{\left( x+1 \right)}}+\frac{C}{{{\left( x+1 \right)}^{2}}}} ∫ ( x − 1 ) ( x + 1 ) 2 2 x 2 − 1 d x = ∫ x − 1 A + ( x + 1 ) B + ( x + 1 ) 2 C 2 x 2 − 1 = A ( x + 1 ) 2 + B ( x + 1 ) ( x + 1 ) + C ( x − 1 ) 2{{x}^{2}}-1=A{{\left( x+1 \right)}^{2}}+B\left( x+1 \right)\left( x+1 \right)+C\left( x-1 \right) 2 x 2 − 1 = A ( x + 1 ) 2 + B ( x + 1 ) ( x + 1 ) + C ( x − 1 ) Put x = 1 ⇒ A = 1 2 x=1\Rightarrow A=\frac{1}{2} x = 1 ⇒ A = 2 1 x = − 1 ⇒ C = − 1 2 x=-1\Rightarrow \,\,C=\frac{-1}{2} x = − 1 ⇒ C = 2 − 1 B = 3 2 B=\frac{3}{2} B = 2 3 I = ∫ 1 2 ( x − 1 ) d x + 3 2 ∫ d x ( x + 1 ) + ( − 1 2 ) ∫ d x ( x − 2 ) 2 I=\int{\frac{\frac{1}{2}}{\left( x-1 \right)}}dx+\frac{3}{2}\int{\frac{dx}{\left( x+1 \right)}}+\left( \frac{-1}{2} \right)\int{\frac{dx}{{{\left( x-2 \right)}^{2}}}} I = ∫ ( x − 1 ) 2 1 d x + 2 3 ∫ ( x + 1 ) d x + ( 2 − 1 ) ∫ ( x − 2 ) 2 d x = 1 2 ℓ n ∣ x − 1 ∣ + 3 2 ( n ∣ x + 1 ) + 1 2 ( x + 2 ) + c =\frac{1}{2}\ell n\left| x-1 \right|+\frac{3}{2}\left( n\left| x+1 \right. \right)+\frac{1}{2\left( x+2 \right)}+c = 2 1 ℓ n ∣ x − 1 ∣ + 2 3 ( n ∣ x + 1 ) + 2 ( x + 2 ) 1 + c ∫ d x ( x + 2 ) ( x 2 + 1 ) = A x + 2 + B x + C x 2 + 1 \int{\frac{dx}{\left( x+2 \right)\left( {{x}^{2}}+1 \right)}}=\frac{A}{x+2}+\frac{Bx+C}{{{x}^{2}}+1} ∫ ( x + 2 ) ( x 2 + 1 ) d x = x + 2 A + x 2 + 1 B x + C
Comparing by x = − 2 ⇒ A = 1 5 x=-2\Rightarrow \,\,A=\frac{1}{5} x = − 2 ⇒ A = 5 1 x 2 {{x}^{2}} x 2 B = − 1 2 B=\frac{-1}{2} B = 2 − 1 C = 2 5 C=\frac{2}{5} C = 5 2 I = 1 5 ∣ d x x + 2 − 1 5 ∫ x x 2 + 1 d x I=\frac{1}{5}\left| \frac{dx}{x+2} \right.-\frac{1}{5}\int{\frac{x}{{{x}^{2}}+1}}\,dx I = 5 1 ∣ ∣ ∣ x + 2 d x − 5 1 ∫ x 2 + 1 x d x I = 1 5 ℓ n ∣ x + 2 ∣ − 1 10 ℓ n ( x 2 + 1 ) + 2 5 tan − 1 ( x ) I=\frac{1}{5}\ell n\left| x+2 \right|-\frac{1}{10}\ell n\left( {{x}^{2}}+1 \right)+\frac{2}{5}{{\tan }^{-1}}\left( x \right) I = 5 1 ℓ n ∣ x + 2 ∣ − 1 0 1 ℓ n ( x 2 + 1 ) + 5 2 tan − 1 ( x )
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
Let us understand with the help of an example:
∫ cos 3 x sin 6 x d x = ∫ 1 − t 2 t 6 d t \int{\frac{{{\cos }^{3}}x}{{{\sin }^{6}}x}dx=\int{\frac{1-{{t}^{2}}}{{{t}^{6}}}dt}} ∫ 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 Type II
∫ d x a cos 2 x + b sin 2 x = ∫ d x a + b sin 2 x \int{\frac{dx}{a{{\cos }^{2}}x+b{{\sin }^{2}}x}}=\int{\frac{dx}{a+b{{\sin }^{2}}x}} ∫ a c o s 2 x + b s i n 2 x d x = ∫ a + b s i n 2 x d x ∫ 1 a + b cos 2 x d x , ∫ 1 ( a sin x + b cos x ) 2 d x \int{\frac{1}{a+b{{\cos }^{2}}x}}\,dx,\int{\frac{1}{{{\left( a\sin x+b\cos x \right)}^{2}}}}\,dx ∫ a + b c o s 2 x 1 d x , ∫ ( a s i n x + b c o s x ) 2 1 d x ∫ 1 a + b sin 2 x + cos 2 x d x \int{\frac{1}{a+b{{\sin }^{2}}x+{{\cos }^{2}}x}}\,dx ∫ a + b s i n 2 x + c o s 2 x 1 d x Rule:
Divide both numerator & denominator by cos 2 x {{\cos }^{2}}x cos 2 x
Illustration:
∫ 1 3 − 4 sin 2 x d x = ∫ 1 cos 2 x 3 cos 2 x − 4 sin 2 x cos 2 x \int{\frac{1}{3-4{{\sin }^{2}}x}}\,dx=\int{\frac{\frac{1}{{{\cos }^{2}}x}}{\frac{3}{{{\cos }^{2}}x}-\frac{4{{\sin }^{2}}x}{{{\cos }^{2}}x}}} ∫ 3 − 4 s i n 2 x 1 d x = ∫ c o s 2 x 3 − c o s 2 x 4 s i n 2 x c o s 2 x 1 = ∫ sec 2 x 3 sec 2 x − 4 tan 2 x d x =\int{\frac{{{\sec }^{2}}x}{3{{\sec }^{2}}x-4{{\tan }^{2}}x}}\,dx = ∫ 3 s e c 2 x − 4 t a n 2 x s e c 2 x d x = ∫ d t 3 ( 1 + t 2 ) − 4 t 2 =\int{\frac{dt}{3\left( 1+{{t}^{2}} \right)-4{{t}^{2}}}} = ∫ 3 ( 1 + t 2 ) − 4 t 2 d t tan x = t sec 2 x d x = d t \tan x=t\,\,\,{{\sec }^{2}}x\,dx=dt tan x = t sec 2 x d x = d t
= ∫ d t 3 − t 2 =\int{\frac{dt}{3-{{t}^{2}}}} = ∫ 3 − t 2 d t = ∫ 1 ( 3 ) 2 − t 2 =\int{\frac{1}{{{\left( \sqrt{3} \right)}^{2}}-{{t}^{2}}}} = ∫ ( 3 ) 2 − t 2 1 = 1 2 3 ℓ n ∣ 3 + t 3 − t ∣ =\frac{1}{2\sqrt{3}}\ell n\left| \frac{\sqrt{3}+t}{\sqrt{3}-t} \right| = 2 3 1 ℓ n ∣ ∣ ∣ ∣ 3 − t 3 + t ∣ ∣ ∣ ∣ = 1 2 3 log ∣ 3 + tan x 3 − tan x ∣ =\frac{1}{2\sqrt{3}}\log \left| \frac{\sqrt{3}+\tan x}{\sqrt{3}-\tan x} \right| = 2 3 1 log ∣ ∣ ∣ ∣ 3 − t a n x 3 + t a n x ∣ ∣ ∣ ∣ Definite Integrals
Choose n divide [a, b] in n parts of width h = b − a n = h=\frac{b-a}{n}= h = n b − a =
∫ a b f ( x ) d x = lim h → 0 h ∑ n = 0 n − 1 f ( a + r h ) \int\limits_{a}^{b}{f\left( x \right)}dx=\underset{h\to 0}{\mathop{\lim }}\,h\,\sum\limits_{n=0}^{n-1}{f\left( a+rh \right)} a ∫ b f ( x ) d x = h → 0 lim h n = 0 ∑ n − 1 f ( a + r h ) Definition of Definite Integral as limit of a solution = lim h → 0 h [ f ( a ) + f ( a + h ) + f ( a + 2 h ) + … . . + f ( a + ( n − 1 ) h ) ] =\underset{h\to 0}{\mathop{\lim }}\,h\left[ f\left( a \right)+f\left( a+h \right)+f\left( a+2h \right)+…..+f\left( a+\left( n-1 \right)h \right) \right] = h → 0 lim h [ f ( a ) + f ( a + h ) + f ( a + 2 h ) + … . . + f ( a + ( n − 1 ) h ) ] n h = b − a nh=b-a n h = b − a Integrals using limits
Evaluate as Limit of a Sum I = ∫ a b x d x ⇒ f ( x ) = x h = b − a n I=\int\limits_{a}^{b}{x\,\,dx\Rightarrow f\left( x \right)=x\,\,\,\,h=\frac{b-a}{n}} I = a ∫ b x d x ⇒ f ( x ) = x h = n b − a f ( a ) = a f\left( a \right)=a f ( a ) = a f ( a + h ) = a + h f\left( a+h \right)=a+h f ( a + h ) = a + h I = h [ a + ( a + h ) + … . ( a + ( n − 1 ) h ) ] I=h\left[ a+\left( a+h \right)+….\left( a+\left( n-1 \right)h \right) \right] I = h [ a + ( a + h ) + … . ( a + ( n − 1 ) h ) ]
= h [ ( n a ) + h + 2 h + 3 h … . ( n − 1 ) h ] =h\left[ \left( na \right)+h+2h+3h….\left( n-1 \right)h \right] = h [ ( n a ) + h + 2 h + 3 h … . ( n − 1 ) h ] = h n a + h 2 n ( n − 1 ) 2 =hna+\frac{{{h}^{2}}n\left( n-1 \right)}{2} = h n a + 2 h 2 n ( n − 1 ) = ( b − a ) a + h 2 n 2 2 − ( b − a ) 2 2 = b 2 − a 2 2 =\left( b-a \right)a+\frac{{{h}^{2}}{{n}^{2}}}{2}-\frac{{{\left( b-a \right)}^{2}}}{2}=\frac{{{b}^{2}}-{{a}^{2}}}{2} = ( b − a ) a + 2 h 2 n 2 − 2 ( b − a ) 2 = 2 b 2 − a 2 ∑ r = 1 n n = n ( n + 1 ) 2 \sum\limits_{r=1}^{n}{n}=\frac{n\left( n+1 \right)}{2} r = 1 ∑ n n = 2 n ( n + 1 ) ∑ r = 1 n n 2 = n ( n + 1 ) ( 2 n + 1 ) 6 \sum\limits_{r=1}^{n}{{{n}^{2}}}=\frac{n\left( n+1 \right)\left( 2n+1 \right)}{6} r = 1 ∑ n n 2 = 6 n ( n + 1 ) ( 2 n + 1 ) ∑ r = 1 n n 3 = [ n ( n + 1 ) 2 ] 2 \sum\limits_{r=1}^{n}{{{n}^{3}}}={{\left[ \frac{n\left( n+1 \right)}{2} \right]}^{2}} r = 1 ∑ n n 3 = [ 2 n ( n + 1 ) ] 2 GP … a + a r + a r 2 … . a r n − 1 = a ( 1 − r n ) 1 − r a+ar+a{{r}^{2}}….a{{r}^{n-1}}=\frac{a\left( 1-{{r}^{n}} \right)}{1-r} a + a r + a r 2 … . a r n − 1 = 1 − r a ( 1 − r n ) sin α + sin ( α + β ) + sin ( α + 2 β ) … . sin ( α + ( n − 1 ) β ) \sin \alpha +\sin \left( \alpha +\beta \right)+\sin \left( \alpha +2\beta \right)….\sin \left( \alpha +\left( n-1 \right)\beta \right) sin α + sin ( α + β ) + sin ( α + 2 β ) … . sin ( α + ( n − 1 ) β ) = sin ( n β 2 ) sin ( β 2 ) sin [ α + α + ( n − 1 ) β 2 ] =\frac{\sin \left( n\frac{\beta }{2} \right)}{\sin \left( \frac{\beta }{2} \right)}\sin \left[ \frac{\alpha +\alpha +\left( n-1 \right)\beta }{2} \right] = s i n ( 2 β ) s i n ( n 2 β ) sin [ 2 α + α + ( n − 1 ) β ] = sin ( n d i f f 2 ) sin ( d i f f 2 ) sin ( 1 s t + l a s t 2 ) =\frac{\sin \left( n\frac{diff}{2} \right)}{\sin \left( \frac{diff}{2} \right)}\sin \left( \frac{1st+last}{2} \right) = s i n ( 2 d i f f ) s i n ( n 2 d i f f ) sin ( 2 1 s t + l a s t ) = sin ( n d i f f 2 ) sin ( d i f f 2 ) cos ( 1 s t + l a s t 2 ) =\frac{\sin \left( \frac{n\,diff}{2} \right)}{\sin \left( \frac{diff}{2} \right)}\cos \left( \frac{1st+last}{2} \right) = s i n ( 2 d i f f ) s i n ( 2 n d i f f ) cos ( 2 1 s t + l a s t ) 1 − 1 2 2 + 1 3 2 − 1 4 2 + 1 5 2 − 1 6 2 = π 2 12 1-\frac{1}{{{2}^{2}}}+\frac{1}{{{3}^{2}}}-\frac{1}{{{4}^{2}}}+\frac{1}{{{5}^{2}}}-\frac{1}{{{6}^{2}}}=\frac{{{\pi }^{2}}}{12} 1 − 2 2 1 + 3 2 1 − 4 2 1 + 5 2 1 − 6 2 1 = 1 2 π 2 1 + 1 2 2 + 1 3 2 + 1 4 2 + 1 5 2 … . = π 2 12 1+\frac{1}{{{2}^{2}}}+\frac{1}{{{3}^{2}}}+\frac{1}{{{4}^{2}}}+\frac{1}{{{5}^{2}}}….=\frac{{{\pi }^{2}}}{12} 1 + 2 2 1 + 3 2 1 + 4 2 1 + 5 2 1 … . = 1 2 π 2 ∫ a b e x d x = lim n → ∞ h [ e a + e a + h + e a + 2 h … e a + ( n − 1 ) h ] \int\limits_{a}^{b}{{{e}^{x}}dx=\underset{n\to \infty }{\mathop{\lim }}\,}h\left[ {{e}^{a}}+{{e}^{a+h}}+{{e}^{a+2h}}…{{e}^{a+\left( n-1 \right)h}} \right] a ∫ b e x d x = n → ∞ lim h [ e a + e a + h + e a + 2 h … e a + ( n − 1 ) h ] = h e a [ 1 + e n + e 2 h + … . ] =h{{e}^{a}}\left[ 1+{{e}^{n}}+{{e}^{2h}}+…. \right] = h e a [ 1 + e n + e 2 h + … . ] = e a [ e n h − 1 ] h ( e n − 1 ) ={{e}^{a}}\left[ {{e}^{nh}}-1 \right]\frac{h}{\left( {{e}^{n}}-1 \right)} = e a [ e n h − 1 ] ( e n − 1 ) h = e b − e a ={{e}^{b}}-{{e}^{a}} = e b − e a ∫ a b sin x d x = lim n → ∞ h [ sin a + sin ( a + n ) + … sin ( a + ( n − 1 ) h ) ] \int\limits_{a}^{b}{\sin x\,dx=\underset{n\to \infty }{\mathop{\lim }}\,}h\left[ \sin a+\sin \left( a+n \right)+…\sin \left( a+\left( n-1 \right)h \right) \right]
