On this page we consider indefinite integrals. See also Definite Integrals.
Maxima can compute indefinite integrals of many functions. Symbolic Integration describes the algorithm for computing indefinite integrals.
integrate(x/(x^3 + 1), x);
2 x - 1
2 atan(-------)
log(x - x + 1) sqrt(3) log(x + 1)
--------------- + ------------- - ----------
6 sqrt(3) 3
To verify this result, we compute its derivative:
diff(%, x);
2 2 x - 1 1
------------------ + -------------- - ---------
2 2 3 (x + 1)
(2 x - 1) 6 (x - x + 1)
3 (---------- + 1)
3
This is a sum of three rational expressions. Obviously, every term of the integral contributes one rational expression. We use ratsimp to bring all three expressions on a common denominator:
ratsimp(%);
x
------
3
x + 1
Let us look at an example with trigonometric functions and exponentials:
integrate(exp(a*x)*sin(x)*cos(x), x) ;
We obtain:
a x %e (a sin(2 x) - 2 cos(2 x)) (%o1) ------------------------------- 2 2 (a + 4)
To verify the result, we compute the derivative:
(%i2) diff(%, x);
a x a x
a %e (a sin(2 x) - 2 cos(2 x)) %e (4 sin(2 x) + 2 a cos(2 x))
(%o2)--------------------------------- + ---------------------------------
2 2
2 (a + 4) 2 (a + 4)
This answer is a surprise, but simplification gives
ratsimp(%);
a x %e sin(2 x) (%o3) -------------- 2
To get rid of the sinus of the multiple angle, we apply trigexpand:
trigexpand(%);
a x (%o4) %e cos(x) sin(x)
Sometimes Maxima answers the so-called 'noun form' of an integral. This tells you that Maxima was unable to find an integral:
integrate(1/((x-3)^4+1/2), x);
/ [ 1 (%o5) I ------------ dx ] 4 1 / (x - 3) + - 2
Sometimes a change of the integration variable helps:
changevar (%, x - 3 - y ,y ,x);
/ [ 1 (%o6) 2 I -------- dy ] 4 / 2 y + 1
Now we can integrate:
ev (%, integrate);
and obtain.
2 3/4 2 3/4 log(sqrt(2) y + 2 y + 1) log(sqrt(2) y - 2 y + 1) (%o8) 2 (---------------------------- - ---------------------------- 3/4 3/4 4 2 4 2 3/4 3/4 2 sqrt(2) y + 2 2 sqrt(2) y - 2 atan(------------------) atan(------------------) 3/4 3/4 2 2 + ------------------------ + ------------------------) 3/4 3/4 2 2 2 2
Backsubstitution gives us the desired integral.
sfx: %, y=x-3;
2 3/4 log(sqrt(2) (x - 3) + 2 (x - 3) + 1) (%o8) 2 (---------------------------------------- 3/4 4 2 3/4 2 sqrt(2) (x - 3) + 2 atan(------------------------) 2 3/4 3/4 log(sqrt(2) (x - 3) - 2 (x - 3) + 1) 2 - ---------------------------------------- + ------------------------------ 3/4 3/4 4 2 2 2 3/4 2 sqrt(2) (x - 3) - 2 atan(------------------------) 3/4 2 + ------------------------------) 3/4 2 2
The integrand is an everywhere positive function. For such functions we can evaluate the integral at the limits of a definite integral to obtain its value. To compute:
'integrate(1/((x-3)^4+1/2), x,0, 1); 1 / [ 1 (%o17) I ------------ dx ] 4 1 / (x - 3) + - 0 2
we can write:
ratsimp(subst (1, x, sfx) - subst(0, x, sfx));
This expression uses substitution to evaluate the indefinite integral at the limits of the definite integral. We obtain:
3/4 3/4
1/4 2 + 6 sqrt(2) 1/4 2 + 4 sqrt(2)
(%o14) (2 2 atan(----------------) - 2 2 atan(----------------)
3/4 3/4
2 2
1/4 3/4 1/4 3/4
+ 2 log(3 2 + 9 sqrt(2) + 1) - 2 log(2 2 + 4 sqrt(2) + 1)
1/4 3/4 1/4 3/4
+ 2 log(- 2 2 + 4 sqrt(2) + 1) - 2 log(- 3 2 + 9 sqrt(2) + 1)
3/4 3/4
1/4 6 sqrt(2) - 2 1/4 4 sqrt(2) - 2
+ 2 2 atan(----------------) - 2 2 atan(----------------))/4
3/4 3/4
2 2
The numerical value is:
(%i15) float(%); (%o15) 0.02880633385274
Numerical integration with the Romberg method confirms that result:
(%i16) romberg(1/((x-3)^4+1/2), x,0, 1); (%o16) 0.02880633392455
The following example shows how an integrand of the form f(x)*f'(x) is handled:
p: (x^3 + x^2)^2*(3*x^2 + 2*x);
2 3 2 2 (%o4) (3 x + 2 x) (x + x )
p2: expand(p);
8 7 6 5 (%o5) 3 x + 8 x + 7 x + 2 x
integrate(p2*sin(2*x^3 + 2*x^2)*cos(x^3 + x^2),x);
We obtain:
3 2 3 2 6 5 4 (%o6)((6 x + 6 x ) sin(3 x + 3 x ) + (- 9 x - 18 x - 9 x + 2) 3 2 3 2 3 2 cos(3 x + 3 x ) + (54 x + 54 x ) sin(x + x ) 6 5 4 3 2 + (- 27 x - 54 x - 27 x + 54) cos(x + x ))/54
The following integral can be computed
integrate((-4*x^3*(x^4 + 1)^(2/3)
-16*x^3*(x^4 + 1)^(1/3))
/(3*x^8 +6*x^4 + 3), x);
We obtain:
1 2 (%o2) ----------- + ----------- 4 1/3 4 2/3 (x + 1) (x + 1)
Simplification with ratsimp gives:
4 2/3 4 1/3 (x + 1) + 2 (x + 1) (%o3) --------------------------- 4 x + 1
A sightly more complicated integral is solved in terms of exponentials and logarithms:
integrate(((6*x^5 + 7*x^4 - 36*x^3 + 18*x - 21)*(x^4 + 1)^(2/3)
+ (2*x^6 - 20*x^4 - 40*x^3 + 18*x^2 + 12)*(x^4 + 1)^(1/3))
/(3*x^8 +6*x^4 + 3), x);
4 4
2 log(x + 1) log(x + 1)
------------- -----------
2 3 3 3
(3 x - 7 x + 9) %e + (2 x + 4 x + 5) %e
(%o4) -----------------------------------------------------------------
4
x + 1
We have to use radcan to obtain a canonical representation of the radicals:
radcan(%);
2 4 2/3 3 4 1/3 (3 x - 7 x + 9) (x + 1) + (2 x + 4 x + 5) (x + 1) (%o5) ----------------------------------------------------------- 4 x + 1
**
(%i6) integrate (7*tan(x)^8 + 7*tan(x)^6, x); 7 5 3 15 tan (x) - 21 tan (x) + 35 tan (x) - 105 tan(x) (%o6) 7 (------------------------------------------------- + x) 105 5 3 3 tan (x) - 5 tan (x) + 15 tan(x) + 7 (--------------------------------- - x) 15 (%i7) ratsimp(%); 7 (%o7) tan (x)
a definite integral: