Integration technique using recurrence relations
In integral calculus, integration by reduction formulae is a method relying on recurrence relations . It is used when an expression containing an integer parameter , usually in the form of powers of elementary functions, or products of transcendental functions and polynomials of arbitrary degree , can't be integrated directly. But using other methods of integration a reduction formula can be set up to obtain the integral of the same or similar expression with a lower integer parameter, progressively simplifying the integral until it can be evaluated. [ 1]
The reduction formula can be derived using any of the common methods of integration, like integration by substitution , integration by parts , integration by trigonometric substitution , integration by partial fractions , etc. The main idea is to express an integral involving an integer parameter (e.g. power) of a function, represented by In , in terms of an integral that involves a lower value of the parameter (lower power) of that function, for example I n -1I n -2recurrence relation . In other words, the reduction formula expresses the integral
I n = ∫ f ( x , n ) d x , {\displaystyle I_{n}=\int f(x,n)\,{\text{d}x,} in terms of
I k = ∫ f ( x , k ) d x , {\displaystyle I_{k}=\int f(x,k)\,{\text{d}x,} where
k < n . {\displaystyle k<n.}
How to compute the integral To compute the integral, we set n to its value and use the reduction formula to express it in terms of the (n – 1) or (n – 2) integral. The lower index integral can be used to calculate the higher index ones; the process is continued repeatedly until we reach a point where the function to be integrated can be computed, usually when its index is 0 or 1. Then we back-substitute the previous results until we have computed In . [ 2]
Examples Below are examples of the procedure.
Cosine integral Typically, integrals like
∫ cos n x d x , {\displaystyle \int \cos ^{n}x\,{\text{d}x,\,\!} can be evaluated by a reduction formula.
∫ cos n ( x ) d x {\displaystyle \int \cos ^{n}(x)\,{\text{d}x\!} n = 1, 2 ... 30Start by setting:
I n = ∫ cos n x d x . {\displaystyle I_{n}=\int \cos ^{n}x\,{\text{d}x.\,\!} Now re-write as:
I n = ∫ cos n − 1 x cos x d x , {\displaystyle I_{n}=\int \cos ^{n-1}x\cos x\,{\text{d}x,\,\!} Integrating by this substitution:
cos x d x = d ( sin x ) , {\displaystyle \cos x\,{\text{d}x={\text{d}(\sin x),\,\!} I n = ∫ cos n − 1 x d ( sin x ) . {\displaystyle I_{n}=\int \cos ^{n-1}x\,{\text{d}(\sin x).\!} Now integrating by parts:
∫ cos n x d x = ∫ cos n − 1 x d ( sin x ) = cos n − 1 x sin x − ∫ sin x d ( cos n − 1 x ) = cos n − 1 x sin x + ( n − 1 ) ∫ sin x cos n − 2 x sin x d x = cos n − 1 x sin x + ( n − 1 ) ∫ cos n − 2 x sin 2 x d x = cos n − 1 x sin x + ( n − 1 ) ∫ cos n − 2 x ( 1 − cos 2 x ) d x = cos n − 1 x sin x + ( n − 1 ) ∫ cos n − 2 x d x − ( n − 1 ) ∫ cos n x d x = cos n − 1 x sin x + ( n − 1 ) I n − 2 − ( n − 1 ) I n , {\displaystyle {\begin{aligned}\int \cos ^{n}x\,{\text{d}x&=\int \cos ^{n-1}x\,{\text{d}(\sin x)\!=\cos ^{n-1}x\sin x-\int \sin x\,{\text{d}(\cos ^{n-1}x)\\&=\cos ^{n-1}x\sin x+(n-1)\int \sin x\cos ^{n-2}x\sin x\,{\text{d}x\\&=\cos ^{n-1}x\sin x+(n-1)\int \cos ^{n-2}x\sin ^{2}x\,{\text{d}x\\&=\cos ^{n-1}x\sin x+(n-1)\int \cos ^{n-2}x(1-\cos ^{2}x)\,{\text{d}x\\&=\cos ^{n-1}x\sin x+(n-1)\int \cos ^{n-2}x\,{\text{d}x-(n-1)\int \cos ^{n}x\,{\text{d}x\\&=\cos ^{n-1}x\sin x+(n-1)I_{n-2}-(n-1)I_{n},\end{aligned}\,} solving for In :
I n + ( n − 1 ) I n = cos n − 1 x sin x + ( n − 1 ) I n − 2 , {\displaystyle I_{n}\ +(n-1)I_{n}\ =\cos ^{n-1}x\sin x\ +\ (n-1)I_{n-2},\,} n I n = cos n − 1 ( x ) sin x + ( n − 1 ) I n − 2 , {\displaystyle nI_{n}\ =\cos ^{n-1}(x)\sin x\ +(n-1)I_{n-2},\,} I n = 1 n cos n − 1 x sin x + n − 1 n I n − 2 , {\displaystyle I_{n}\ ={\frac {1}{n}\cos ^{n-1}x\sin x\ +{\frac {n-1}{n}I_{n-2},\,} so the reduction formula is:
∫ cos n x d x = 1 n cos n − 1 x sin x + n − 1 n ∫ cos n − 2 x d x . {\displaystyle \int \cos ^{n}x\,{\text{d}x\ ={\frac {1}{n}\cos ^{n-1}x\sin x+{\frac {n-1}{n}\int \cos ^{n-2}x\,{\text{d}x.\!} To supplement the example, the above can be used to evaluate the integral for (say) n = 5;
I 5 = ∫ cos 5 x d x . {\displaystyle I_{5}=\int \cos ^{5}x\,{\text{d}x.\,\!} Calculating lower indices:
n = 5 , I 5 = 1 5 cos 4 x sin x + 4 5 I 3 , {\displaystyle n=5,\quad I_{5}={\tfrac {1}{5}\cos ^{4}x\sin x+{\tfrac {4}{5}I_{3},\,} n = 3 , I 3 = 1 3 cos 2 x sin x + 2 3 I 1 , {\displaystyle n=3,\quad I_{3}={\tfrac {1}{3}\cos ^{2}x\sin x+{\tfrac {2}{3}I_{1},\,} back-substituting:
∵ I 1 = ∫ cos x d x = sin x + C 1 , {\displaystyle \because I_{1}\ =\int \cos x\,{\text{d}x=\sin x+C_{1},\,} ∴ I 3 = 1 3 cos 2 x sin x + 2 3 sin x + C 2 , C 2 = 2 3 C 1 , {\displaystyle \therefore I_{3}\ ={\tfrac {1}{3}\cos ^{2}x\sin x+{\tfrac {2}{3}\sin x+C_{2},\quad C_{2}\ ={\tfrac {2}{3}C_{1},\,} I 5 = 1 5 cos 4 x sin x + 4 5 [ 1 3 cos 2 x sin x + 2 3 sin x ] + C , {\displaystyle I_{5}\ ={\frac {1}{5}\cos ^{4}x\sin x+{\frac {4}{5}\left[{\frac {1}{3}\cos ^{2}x\sin x+{\frac {2}{3}\sin x\right]+C,\,} where C is a constant.
Exponential integral Another typical example is:
∫ x n e a x d x . {\displaystyle \int x^{n}e^{ax}\,{\text{d}x.\,\!} Start by setting:
I n = ∫ x n e a x d x . {\displaystyle I_{n}=\int x^{n}e^{ax}\,{\text{d}x.\,\!} Integrating by substitution:
x n d x = d ( x n + 1 ) n + 1 , {\displaystyle x^{n}\,{\text{d}x={\frac {\text{d}(x^{n+1})}{n+1},\,\!} I n = 1 n + 1 ∫ e a x d ( x n + 1 ) , {\displaystyle I_{n}={\frac {1}{n+1}\int e^{ax}\,{\text{d}(x^{n+1}),\!} Now integrating by parts:
∫ e a x d ( x n + 1 ) = x n + 1 e a x − ∫ x n + 1 d ( e a x ) = x n + 1 e a x − a ∫ x n + 1 e a x d x , {\displaystyle {\begin{aligned}\int e^{ax}\,{\text{d}(x^{n+1})&=x^{n+1}e^{ax}-\int x^{n+1}\,{\text{d}(e^{ax})\\&=x^{n+1}e^{ax}-a\int x^{n+1}e^{ax}\,{\text{d}x,\end{aligned}\!} ( n + 1 ) I n = x n + 1 e a x − a I n + 1 , {\displaystyle (n+1)I_{n}=x^{n+1}e^{ax}-aI_{n+1},\!} shifting indices back by 1 (so n + 1 → n , n → n – 1):
n I n − 1 = x n e a x − a I n , {\displaystyle nI_{n-1}=x^{n}e^{ax}-aI_{n},\!} solving for In :
I n = 1 a ( x n e a x − n I n − 1 ) , {\displaystyle I_{n}={\frac {1}{a}\left(x^{n}e^{ax}-nI_{n-1}\right),\,\!} so the reduction formula is:
∫ x n e a x d x = 1 a ( x n e a x − n ∫ x n − 1 e a x d x ) . {\displaystyle \int x^{n}e^{ax}\,{\text{d}x={\frac {1}{a}\left(x^{n}e^{ax}-n\int x^{n-1}e^{ax}\,{\text{d}x\right).\!} An alternative way in which the derivation could be done starts by substituting e a x {\displaystyle e^{ax}
Integration by substitution:
e a x d x = d ( e a x ) a , {\displaystyle e^{ax}\,{\text{d}x={\frac {\text{d}(e^{ax})}{a},\,\!}
I n = 1 a ∫ x n d ( e a x ) , {\displaystyle I_{n}={\frac {1}{a}\int x^{n}\,{\text{d}(e^{ax}),\!}
Now integrating by parts:
∫ x n d ( e a x ) = x n e a x − ∫ e a x d ( x n ) = x n e a x − n ∫ e a x x n − 1 d x , {\displaystyle {\begin{aligned}\int x^{n}\,{\text{d}(e^{ax})&=x^{n}e^{ax}-\int e^{ax}\,{\text{d}(x^{n})\\&=x^{n}e^{ax}-n\int e^{ax}x^{n-1}\,{\text{d}x,\end{aligned}\!}
which gives the reduction formula when substituting back:
I n = 1 a ( x n e a x − n I n − 1 ) , {\displaystyle I_{n}={\frac {1}{a}\left(x^{n}e^{ax}-nI_{n-1}\right),\,\!}
which is equivalent to:
∫ x n e a x d x = 1 a ( x n e a x − n ∫ x n − 1 e a x d x ) . {\displaystyle \int x^{n}e^{ax}\,{\text{d}x={\frac {1}{a}\left(x^{n}e^{ax}-n\int x^{n-1}e^{ax}\,{\text{d}x\right).\!} Another alternative way in which the derivation could be done by integrating by parts:
I n = ∫ x n x e a x d x , {\displaystyle I_{n}=\int x^{n}xe^{ax}\,{\text{d}x,\!} u = x n , d v = e a x , {\displaystyle u=x^{n}{\text{ , }\ dv=e^{ax},} d u d x = n x n − 1 , v = e a x a {\displaystyle {\frac {du}{dx}\ =nx^{n-1}{\text{ , }\ v={\frac {e^{ax}{a}\ } I n = x n e a x a − ∫ n x n − 1 e a x a d x {\displaystyle I_{n}={\frac {x^{n}e^{ax}{a}\ -\int nx^{n-1}\ {\frac {e^{ax}{a}\ {\text{d}x\ } I n = x n e a x a − n a ∫ x n − 1 e a x d x {\displaystyle I_{n}={\frac {x^{n}e^{ax}{a}\ -{\frac {n}{a}\ \int x^{n-1}e^{ax}\ {\text{d}x\ } Remember:
I n − 1 = ∫ x n − 1 e a x d x {\displaystyle I_{n-1}=\int x^{n-1}e^{ax}\ {\text{d}x\ } ∴ I n = x n e a x a − n a I n − 1 {\displaystyle \therefore \ I_{n}={\frac {x^{n}e^{ax}{a}\ -{\frac {n}{a}\ I_{n-1} which gives the reduction formula when substituting back:
I n = 1 a ( x n e a x − n I n − 1 ) , {\displaystyle I_{n}={\frac {1}{a}\left(x^{n}e^{ax}-nI_{n-1}\right),\,\!} which is equivalent to:
∫ x n e a x d x = 1 a ( x n e a x − n ∫ x n − 1 e a x d x ) . {\displaystyle \int x^{n}e^{ax}\,{\text{d}x={\frac {1}{a}\left(x^{n}e^{ax}-n\int x^{n-1}e^{ax}\,{\text{d}x\right).\!}
Rational functions The following integrals[ 3]
Factors of the linear radical a x + b {\displaystyle {\sqrt {ax+b}\,\!} Linear factors p x + q {\displaystyle {px+q}\,\!} a x + b {\displaystyle {\sqrt {ax+b}\,\!} Quadratic factors x 2 + a 2 {\displaystyle x^{2}+a^{2}\,\!} Quadratic factors x 2 − a 2 {\displaystyle x^{2}-a^{2}\,\!} x > a {\displaystyle x>a\,\!} Quadratic factors a 2 − x 2 {\displaystyle a^{2}-x^{2}\,\!} x < a {\displaystyle x<a\,\!} (Irreducible ) quadratic factors a x 2 + b x + c {\displaystyle ax^{2}+bx+c\,\!} Radicals of irreducible quadratic factors a x 2 + b x + c {\displaystyle {\sqrt {ax^{2}+bx+c}\,\!} Integral Reduction formula
I n = ∫ x n a x + b d x {\displaystyle I_{n}=\int {\frac {x^{n}{\sqrt {ax+b}\,{\text{d}x\,\!} I n = 2 x n a x + b a ( 2 n + 1 ) − 2 n b a ( 2 n + 1 ) I n − 1 {\displaystyle I_{n}={\frac {2x^{n}{\sqrt {ax+b}{a(2n+1)}-{\frac {2nb}{a(2n+1)}I_{n-1}\,\!} I n = ∫ d x x n a x + b {\displaystyle I_{n}=\int {\frac {\text{d}x}{x^{n}{\sqrt {ax+b}\,\!} I n = − a x + b ( n − 1 ) b x n − 1 − a ( 2 n − 3 ) 2 b ( n − 1 ) I n − 1 {\displaystyle I_{n}=-{\frac {\sqrt {ax+b}{(n-1)bx^{n-1}-{\frac {a(2n-3)}{2b(n-1)}I_{n-1}\,\!} I n = ∫ x n a x + b d x {\displaystyle I_{n}=\int x^{n}{\sqrt {ax+b}\,{\text{d}x\,\!} I n = 2 x n ( a x + b ) 3 a ( 2 n + 3 ) − 2 n b a ( 2 n + 3 ) I n − 1 {\displaystyle I_{n}={\frac {2x^{n}{\sqrt {(ax+b)^{3}{a(2n+3)}-{\frac {2nb}{a(2n+3)}I_{n-1}\,\!} I m , n = ∫ d x ( a x + b ) m ( p x + q ) n {\displaystyle I_{m,n}=\int {\frac {\text{d}x}{(ax+b)^{m}(px+q)^{n}\,\!} I m , n = { − 1 ( n − 1 ) ( b p − a q ) [ 1 ( a x + b ) m − 1 ( p x + q ) n − 1 + a ( m + n − 2 ) I m , n − 1 ] 1 ( m − 1 ) ( b p − a q ) [ 1 ( a x + b ) m − 1 ( p x + q ) n − 1 + p ( m + n − 2 ) I m − 1 , n ] {\displaystyle I_{m,n}={\begin{cases}-{\frac {1}{(n-1)(bp-aq)}\left[{\frac {1}{(ax+b)^{m-1}(px+q)^{n-1}+a(m+n-2)I_{m,n-1}\right]\\{\frac {1}{(m-1)(bp-aq)}\left[{\frac {1}{(ax+b)^{m-1}(px+q)^{n-1}+p(m+n-2)I_{m-1,n}\right]\end{cases}\,\!} I m , n = ∫ ( a x + b ) m ( p x + q ) n d x {\displaystyle I_{m,n}=\int {\frac {(ax+b)^{m}{(px+q)^{n}\,{\text{d}x\,\!} I m , n = { − 1 ( n − 1 ) ( b p − a q ) [ ( a x + b ) m + 1 ( p x + q ) n − 1 + a ( n − m − 2 ) I m , n − 1 ] − 1 ( n − m − 1 ) p [ ( a x + b ) m ( p x + q ) n − 1 + m ( b p − a q ) I m − 1 , n ] − 1 ( n − 1 ) p [ ( a x + b ) m ( p x + q ) n − 1 − a m I m − 1 , n − 1 ] {\displaystyle I_{m,n}={\begin{cases}-{\frac {1}{(n-1)(bp-aq)}\left[{\frac {(ax+b)^{m+1}{(px+q)^{n-1}+a(n-m-2)I_{m,n-1}\right]\\-{\frac {1}{(n-m-1)p}\left[{\frac {(ax+b)^{m}{(px+q)^{n-1}+m(bp-aq)I_{m-1,n}\right]\\-{\frac {1}{(n-1)p}\left[{\frac {(ax+b)^{m}{(px+q)^{n-1}-amI_{m-1,n-1}\right]\end{cases}\,\!}
Integral Reduction formula
I n = ∫ d x ( x 2 + a 2 ) n {\displaystyle I_{n}=\int {\frac {\text{d}x}{(x^{2}+a^{2})^{n}\,\!} I n = x 2 a 2 ( n − 1 ) ( x 2 + a 2 ) n − 1 + 2 n − 3 2 a 2 ( n − 1 ) I n − 1 {\displaystyle I_{n}={\frac {x}{2a^{2}(n-1)(x^{2}+a^{2})^{n-1}+{\frac {2n-3}{2a^{2}(n-1)}I_{n-1}\,\!} I n , m = ∫ d x x m ( x 2 + a 2 ) n {\displaystyle I_{n,m}=\int {\frac {\text{d}x}{x^{m}(x^{2}+a^{2})^{n}\,\!} a 2 I n , m = I m , n − 1 − I m − 2 , n {\displaystyle a^{2}I_{n,m}=I_{m,n-1}-I_{m-2,n}\,\!} I n , m = ∫ x m ( x 2 + a 2 ) n d x {\displaystyle I_{n,m}=\int {\frac {x^{m}{(x^{2}+a^{2})^{n}\,{\text{d}x\,\!} I n , m = I m − 2 , n − 1 − a 2 I m − 2 , n {\displaystyle I_{n,m}=I_{m-2,n-1}-a^{2}I_{m-2,n}\,\!}
Integral Reduction formula
I n = ∫ d x ( x 2 − a 2 ) n {\displaystyle I_{n}=\int {\frac {\text{d}x}{(x^{2}-a^{2})^{n}\,\!} I n = − x 2 a 2 ( n − 1 ) ( x 2 − a 2 ) n − 1 − 2 n − 3 2 a 2 ( n − 1 ) I n − 1 {\displaystyle I_{n}=-{\frac {x}{2a^{2}(n-1)(x^{2}-a^{2})^{n-1}-{\frac {2n-3}{2a^{2}(n-1)}I_{n-1}\,\!} I n , m = ∫ d x x m ( x 2 − a 2 ) n {\displaystyle I_{n,m}=\int {\frac {\text{d}x}{x^{m}(x^{2}-a^{2})^{n}\,\!} a 2 I n , m = I m − 2 , n − I m , n − 1 {\displaystyle {a^{2}I_{n,m}=I_{m-2,n}-I_{m,n-1}\,\!} I n , m = ∫ x m ( x 2 − a 2 ) n d x {\displaystyle I_{n,m}=\int {\frac {x^{m}{(x^{2}-a^{2})^{n}\,{\text{d}x\,\!} I n , m = I m − 2 , n − 1 + a 2 I m − 2 , n {\displaystyle I_{n,m}=I_{m-2,n-1}+a^{2}I_{m-2,n}\,\!}
Integral Reduction formula
I n = ∫ d x ( a 2 − x 2 ) n {\displaystyle I_{n}=\int {\frac {\text{d}x}{(a^{2}-x^{2})^{n}\,\!} I n = x 2 a 2 ( n − 1 ) ( a 2 − x 2 ) n − 1 + 2 n − 3 2 a 2 ( n − 1 ) I n − 1 {\displaystyle I_{n}={\frac {x}{2a^{2}(n-1)(a^{2}-x^{2})^{n-1}+{\frac {2n-3}{2a^{2}(n-1)}I_{n-1}\,\!} I n , m = ∫ d x x m ( a 2 − x 2 ) n {\displaystyle I_{n,m}=\int {\frac {\text{d}x}{x^{m}(a^{2}-x^{2})^{n}\,\!} a 2 I n , m = I m , n − 1 + I m − 2 , n {\displaystyle {a^{2}I_{n,m}=I_{m,n-1}+I_{m-2,n}\,\!} I n , m = ∫ x m ( a 2 − x 2 ) n d x {\displaystyle I_{n,m}=\int {\frac {x^{m}{(a^{2}-x^{2})^{n}\,{\text{d}x\,\!} I n , m = a 2 I m − 2 , n − I m − 2 , n − 1 {\displaystyle I_{n,m}=a^{2}I_{m-2,n}-I_{m-2,n-1}\,\!}
Integral Reduction formula
I n = ∫ d x x n ( a x 2 + b x + c ) {\displaystyle I_{n}=\int {\frac {\text{d}x}{x^{n}(ax^{2}+bx+c)}\,\!} − c I n = 1 x n − 1 ( n − 1 ) + b I n − 1 + a I n − 2 {\displaystyle -cI_{n}={\frac {1}{x^{n-1}(n-1)}+bI_{n-1}+aI_{n-2}\,\!} I m , n = ∫ x m d x ( a x 2 + b x + c ) n {\displaystyle I_{m,n}=\int {\frac {x^{m}\,{\text{d}x}{(ax^{2}+bx+c)^{n}\,\!} I m , n = − x m − 1 a ( 2 n − m − 1 ) ( a x 2 + b x + c ) n − 1 − b ( n − m ) a ( 2 n − m − 1 ) I m − 1 , n + c ( m − 1 ) a ( 2 n − m − 1 ) I m − 2 , n {\displaystyle I_{m,n}=-{\frac {x^{m-1}{a(2n-m-1)(ax^{2}+bx+c)^{n-1}-{\frac {b(n-m)}{a(2n-m-1)}I_{m-1,n}+{\frac {c(m-1)}{a(2n-m-1)}I_{m-2,n}\,\!} I m , n = ∫ d x x m ( a x 2 + b x + c ) n {\displaystyle I_{m,n}=\int {\frac {\text{d}x}{x^{m}(ax^{2}+bx+c)^{n}\,\!} − c ( m − 1 ) I m , n = 1 x m − 1 ( a x 2 + b x + c ) n − 1 + a ( m + 2 n − 3 ) I m − 2 , n + b ( m + n − 2 ) I m − 1 , n {\displaystyle -c(m-1)I_{m,n}={\frac {1}{x^{m-1}(ax^{2}+bx+c)^{n-1}+{a(m+2n-3)}I_{m-2,n}+{b(m+n-2)}I_{m-1,n}\,\!}
Integral Reduction formula
I n = ∫ ( a x 2 + b x + c ) n d x {\displaystyle I_{n}=\int (ax^{2}+bx+c)^{n}\,{\text{d}x\,\!} 8 a ( n + 1 ) I n + 1 2 = 2 ( 2 a x + b ) ( a x 2 + b x + c ) n + 1 2 + ( 2 n + 1 ) ( 4 a c − b 2 ) I n − 1 2 {\displaystyle 8a(n+1)I_{n+{\frac {1}{2}=2(2ax+b)(ax^{2}+bx+c)^{n+{\frac {1}{2}+(2n+1)(4ac-b^{2})I_{n-{\frac {1}{2}\,\!} I n = ∫ 1 ( a x 2 + b x + c ) n d x {\displaystyle I_{n}=\int {\frac {1}{(ax^{2}+bx+c)^{n}\,{\text{d}x\,\!} ( 2 n − 1 ) ( 4 a c − b 2 ) I n + 1 2 = 2 ( 2 a x + b ) ( a x 2 + b x + c ) n − 1 2 + 8 a ( n − 1 ) I n − 1 2 {\displaystyle (2n-1)(4ac-b^{2})I_{n+{\frac {1}{2}={\frac {2(2ax+b)}{(ax^{2}+bx+c)^{n-{\frac {1}{2}+{8a(n-1)}I_{n-{\frac {1}{2}\,\!}
note that by the laws of indices :
I n + 1 2 = I 2 n + 1 2 = ∫ 1 ( a x 2 + b x + c ) 2 n + 1 2 d x = ∫ 1 ( a x 2 + b x + c ) 2 n + 1 d x {\displaystyle I_{n+{\frac {1}{2}=I_{\frac {2n+1}{2}=\int {\frac {1}{(ax^{2}+bx+c)^{\frac {2n+1}{2}\,{\text{d}x=\int {\frac {1}{\sqrt {(ax^{2}+bx+c)^{2n+1}\,{\text{d}x\,\!}
Transcendental functions The following integrals[ 4]
Factors of sine Factors of cosine Factors of sine and cosine products and quotients Products/quotients of exponential factors and powers of x Products of exponential and sine/cosine factors Integral Reduction formula
I n = ∫ x n sin a x d x {\displaystyle I_{n}=\int x^{n}\sin {ax}\,{\text{d}x\,\!} a 2 I n = − a x n cos a x + n x n − 1 sin a x − n ( n − 1 ) I n − 2 {\displaystyle a^{2}I_{n}=-ax^{n}\cos {ax}+nx^{n-1}\sin {ax}-n(n-1)I_{n-2}\,\!} J n = ∫ x n cos a x d x {\displaystyle J_{n}=\int x^{n}\cos {ax}\,{\text{d}x\,\!} a 2 J n = a x n sin a x + n x n − 1 cos a x − n ( n − 1 ) J n − 2 {\displaystyle a^{2}J_{n}=ax^{n}\sin {ax}+nx^{n-1}\cos {ax}-n(n-1)J_{n-2}\,\!} I n = ∫ sin a x x n d x {\displaystyle I_{n}=\int {\frac {\sin {ax}{x^{n}\,{\text{d}x\,\!} J n = ∫ cos a x x n d x {\displaystyle J_{n}=\int {\frac {\cos {ax}{x^{n}\,{\text{d}x\,\!}
I n = − sin a x ( n − 1 ) x n − 1 + a n − 1 J n − 1 {\displaystyle I_{n}=-{\frac {\sin {ax}{(n-1)x^{n-1}+{\frac {a}{n-1}J_{n-1}\,\!} J n = − cos a x ( n − 1 ) x n − 1 − a n − 1 I n − 1 {\displaystyle J_{n}=-{\frac {\cos {ax}{(n-1)x^{n-1}-{\frac {a}{n-1}I_{n-1}\,\!}
the formulae can be combined to obtain separate equations in In :
J n − 1 = − cos a x ( n − 2 ) x n − 2 − a n − 2 I n − 2 {\displaystyle J_{n-1}=-{\frac {\cos {ax}{(n-2)x^{n-2}-{\frac {a}{n-2}I_{n-2}\,\!}
I n = − sin a x ( n − 1 ) x n − 1 − a n − 1 [ cos a x ( n − 2 ) x n − 2 + a n − 2 I n − 2 ] {\displaystyle I_{n}=-{\frac {\sin {ax}{(n-1)x^{n-1}-{\frac {a}{n-1}\left[{\frac {\cos {ax}{(n-2)x^{n-2}+{\frac {a}{n-2}I_{n-2}\right]\,\!}
∴ I n = − sin a x ( n − 1 ) x n − 1 − a ( n − 1 ) ( n − 2 ) ( cos a x x n − 2 + a I n − 2 ) {\displaystyle \therefore I_{n}=-{\frac {\sin {ax}{(n-1)x^{n-1}-{\frac {a}{(n-1)(n-2)}\left({\frac {\cos {ax}{x^{n-2}+aI_{n-2}\right)\,\!}
and Jn :
I n − 1 = − sin a x ( n − 2 ) x n − 2 + a n − 2 J n − 2 {\displaystyle I_{n-1}=-{\frac {\sin {ax}{(n-2)x^{n-2}+{\frac {a}{n-2}J_{n-2}\,\!}
J n = − cos a x ( n − 1 ) x n − 1 − a n − 1 [ − sin a x ( n − 2 ) x n − 2 + a n − 2 J n − 2 ] {\displaystyle J_{n}=-{\frac {\cos {ax}{(n-1)x^{n-1}-{\frac {a}{n-1}\left[-{\frac {\sin {ax}{(n-2)x^{n-2}+{\frac {a}{n-2}J_{n-2}\right]\,\!}
∴ J n = − cos a x ( n − 1 ) x n − 1 − a ( n − 1 ) ( n − 2 ) ( − sin a x x n − 2 + a J n − 2 ) {\displaystyle \therefore J_{n}=-{\frac {\cos {ax}{(n-1)x^{n-1}-{\frac {a}{(n-1)(n-2)}\left(-{\frac {\sin {ax}{x^{n-2}+aJ_{n-2}\right)\,\!}
I n = ∫ sin n a x d x {\displaystyle I_{n}=\int \sin ^{n}{ax}\,{\text{d}x\,\!} a n I n = − sin n − 1 a x cos a x + a ( n − 1 ) I n − 2 {\displaystyle anI_{n}=-\sin ^{n-1}{ax}\cos {ax}+a(n-1)I_{n-2}\,\!} J n = ∫ cos n a x d x {\displaystyle J_{n}=\int \cos ^{n}{ax}\,{\text{d}x\,\!} a n J n = sin a x cos n − 1 a x + a ( n − 1 ) J n − 2 {\displaystyle anJ_{n}=\sin {ax}\cos ^{n-1}{ax}+a(n-1)J_{n-2}\,\!} I n = ∫ d x sin n a x {\displaystyle I_{n}=\int {\frac {\text{d}x}{\sin ^{n}{ax}\,\!} ( n − 1 ) I n = − cos a x a sin n − 1 a x + ( n − 2 ) I n − 2 {\displaystyle (n-1)I_{n}=-{\frac {\cos {ax}{a\sin ^{n-1}{ax}+(n-2)I_{n-2}\,\!} J n = ∫ d x cos n a x {\displaystyle J_{n}=\int {\frac {\text{d}x}{\cos ^{n}{ax}\,\!} ( n − 1 ) J n = sin a x a cos n − 1 a x + ( n − 2 ) J n − 2 {\displaystyle (n-1)J_{n}={\frac {\sin {ax}{a\cos ^{n-1}{ax}+(n-2)J_{n-2}\,\!}
Integral Reduction formula
I m , n = ∫ sin m a x cos n a x d x {\displaystyle I_{m,n}=\int \sin ^{m}{ax}\cos ^{n}{ax}\,{\text{d}x\,\!} I m , n = { − sin m − 1 a x cos n + 1 a x a ( m + n ) + m − 1 m + n I m − 2 , n sin m + 1 a x cos n − 1 a x a ( m + n ) + n − 1 m + n I m , n − 2 {\displaystyle I_{m,n}={\begin{cases}-{\frac {\sin ^{m-1}{ax}\cos ^{n+1}{ax}{a(m+n)}+{\frac {m-1}{m+n}I_{m-2,n}\\{\frac {\sin ^{m+1}{ax}\cos ^{n-1}{ax}{a(m+n)}+{\frac {n-1}{m+n}I_{m,n-2}\\\end{cases}\,\!} I m , n = ∫ d x sin m a x cos n a x {\displaystyle I_{m,n}=\int {\frac {\text{d}x}{\sin ^{m}{ax}\cos ^{n}{ax}\,\!} I m , n = { 1 a ( n − 1 ) sin m − 1 a x cos n − 1 a x + m + n − 2 n − 1 I m , n − 2 − 1 a ( m − 1 ) sin m − 1 a x cos n − 1 a x + m + n − 2 m − 1 I m − 2 , n {\displaystyle I_{m,n}={\begin{cases}{\frac {1}{a(n-1)\sin ^{m-1}{ax}\cos ^{n-1}{ax}+{\frac {m+n-2}{n-1}I_{m,n-2}\\-{\frac {1}{a(m-1)\sin ^{m-1}{ax}\cos ^{n-1}{ax}+{\frac {m+n-2}{m-1}I_{m-2,n}\\\end{cases}\,\!} I m , n = ∫ sin m a x cos n a x d x {\displaystyle I_{m,n}=\int {\frac {\sin ^{m}{ax}{\cos ^{n}{ax}\,{\text{d}x\,\!} I m , n = { sin m − 1 a x a ( n − 1 ) cos n − 1 a x − m − 1 n − 1 I m − 2 , n − 2 sin m + 1 a x a ( n − 1 ) cos n − 1 a x − m − n + 2 n − 1 I m , n − 2 − sin m − 1 a x a ( m − n ) cos n − 1 a x + m − 1 m − n I m − 2 , n {\displaystyle I_{m,n}={\begin{cases}{\frac {\sin ^{m-1}{ax}{a(n-1)\cos ^{n-1}{ax}-{\frac {m-1}{n-1}I_{m-2,n-2}\\{\frac {\sin ^{m+1}{ax}{a(n-1)\cos ^{n-1}{ax}-{\frac {m-n+2}{n-1}I_{m,n-2}\\-{\frac {\sin ^{m-1}{ax}{a(m-n)\cos ^{n-1}{ax}+{\frac {m-1}{m-n}I_{m-2,n}\\\end{cases}\,\!} I m , n = ∫ cos m a x sin n a x d x {\displaystyle I_{m,n}=\int {\frac {\cos ^{m}{ax}{\sin ^{n}{ax}\,{\text{d}x\,\!} I m , n = { − cos m − 1 a x a ( n − 1 ) sin n − 1 a x − m − 1 n − 1 I m − 2 , n − 2 − cos m + 1 a x a ( n − 1 ) sin n − 1 a x − m − n + 2 n − 1 I m , n − 2 cos m − 1 a x a ( m − n ) sin n − 1 a x + m − 1 m − n I m − 2 , n {\displaystyle I_{m,n}={\begin{cases}-{\frac {\cos ^{m-1}{ax}{a(n-1)\sin ^{n-1}{ax}-{\frac {m-1}{n-1}I_{m-2,n-2}\\-{\frac {\cos ^{m+1}{ax}{a(n-1)\sin ^{n-1}{ax}-{\frac {m-n+2}{n-1}I_{m,n-2}\\{\frac {\cos ^{m-1}{ax}{a(m-n)\sin ^{n-1}{ax}+{\frac {m-1}{m-n}I_{m-2,n}\\\end{cases}\,\!}
Integral Reduction formula
I n = ∫ x n e a x d x {\displaystyle I_{n}=\int x^{n}e^{ax}\,{\text{d}x\,\!} n > 0 {\displaystyle n>0\,\!}
I n = x n e a x a − n a I n − 1 {\displaystyle I_{n}={\frac {x^{n}e^{ax}{a}-{\frac {n}{a}I_{n-1}\,\!} I n = ∫ x − n e a x d x {\displaystyle I_{n}=\int x^{-n}e^{ax}\,{\text{d}x\,\!} n > 0 {\displaystyle n>0\,\!}
n ≠ 1 {\displaystyle n\neq 1\,\!}
I n = − e a x ( n − 1 ) x n − 1 + a n − 1 I n − 1 {\displaystyle I_{n}={\frac {-e^{ax}{(n-1)x^{n-1}+{\frac {a}{n-1}I_{n-1}\,\!} I n = ∫ e a x sin n b x d x {\displaystyle I_{n}=\int e^{ax}\sin ^{n}{bx}\,{\text{d}x\,\!} I n = e a x sin n − 1 b x a 2 + ( b n ) 2 ( a sin b x − b n cos b x ) + n ( n − 1 ) b 2 a 2 + ( b n ) 2 I n − 2 {\displaystyle I_{n}={\frac {e^{ax}\sin ^{n-1}{bx}{a^{2}+(bn)^{2}\left(a\sin bx-bn\cos bx\right)+{\frac {n(n-1)b^{2}{a^{2}+(bn)^{2}I_{n-2}\,\!} I n = ∫ e a x cos n b x d x {\displaystyle I_{n}=\int e^{ax}\cos ^{n}{bx}\,{\text{d}x\,\!} I n = e a x cos n − 1 b x a 2 + ( b n ) 2 ( a cos b x + b n sin b x ) + n ( n − 1 ) b 2 a 2 + ( b n ) 2 I n − 2 {\displaystyle I_{n}={\frac {e^{ax}\cos ^{n-1}{bx}{a^{2}+(bn)^{2}\left(a\cos bx+bn\sin bx\right)+{\frac {n(n-1)b^{2}{a^{2}+(bn)^{2}I_{n-2}\,\!}
References ^ Mathematical methods for physics and engineering, K.F. Riley, M.P. Hobson, S.J. Bence, Cambridge University Press, 2010, ISBN 978-0-521-86153-3 ^ Further Elementary Analysis, R.I. Porter, G. Bell & Sons Ltd, 1978, ISBN 0-7135-1594-5 ^ http://www.sosmath.com/tables/tables.html -> Indefinite integrals list^ http://www.sosmath.com/tables/tables.html -> Indefinite integrals list
Bibliography Anton, Bivens, Davis, Calculus, 7th edition.
![{\displaystyle I_{5}\ ={\frac {1}{5}\cos ^{4}x\sin x+{\frac {4}{5}\left[{\frac {1}{3}\cos ^{2}x\sin x+{\frac {2}{3}\sin x\right]+C,\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7803bf2d86071f40537d4590e76ab498354b7c7f)
![{\displaystyle I_{m,n}={\begin{cases}-{\frac {1}{(n-1)(bp-aq)}\left[{\frac {1}{(ax+b)^{m-1}(px+q)^{n-1}+a(m+n-2)I_{m,n-1}\right]\\{\frac {1}{(m-1)(bp-aq)}\left[{\frac {1}{(ax+b)^{m-1}(px+q)^{n-1}+p(m+n-2)I_{m-1,n}\right]\end{cases}\,\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/22e62922721540ee6636beadc5183d305fa0a98b)
![{\displaystyle I_{m,n}={\begin{cases}-{\frac {1}{(n-1)(bp-aq)}\left[{\frac {(ax+b)^{m+1}{(px+q)^{n-1}+a(n-m-2)I_{m,n-1}\right]\\-{\frac {1}{(n-m-1)p}\left[{\frac {(ax+b)^{m}{(px+q)^{n-1}+m(bp-aq)I_{m-1,n}\right]\\-{\frac {1}{(n-1)p}\left[{\frac {(ax+b)^{m}{(px+q)^{n-1}-amI_{m-1,n-1}\right]\end{cases}\,\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/18cf63aea753eb3517e589623b4728e1545c012b)
![{\displaystyle I_{n}=-{\frac {\sin {ax}{(n-1)x^{n-1}-{\frac {a}{n-1}\left[{\frac {\cos {ax}{(n-2)x^{n-2}+{\frac {a}{n-2}I_{n-2}\right]\,\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c59ab79c10f4eed2620c496fac8b476174a84477)
![{\displaystyle J_{n}=-{\frac {\cos {ax}{(n-1)x^{n-1}-{\frac {a}{n-1}\left[-{\frac {\sin {ax}{(n-2)x^{n-2}+{\frac {a}{n-2}J_{n-2}\right]\,\!}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cf2ed60b99d5d7b4dfa736c3627c68021c847b44)
