Antiderivatives and definite integrals

### 13.3.1 Antiderivatives and definite integrals

The int and integrate commands compute a primitive or a definite integral. A difference between the two commands is that if you input quest() just after the evaluation of integrate, the answer is written with the ∫ symbol.

Int is the inert form of integrate; namely, it evaluates to integrate for later evaluation.

- To find a primitive (an antiderivative), int (or integrate) takes one mandatory argument and one optional argument:
  - expr, an expression.
  - Optionally, x, the name of a variable (by default the value is x, so if the variable is x the second argument is unnecessary).
- int(expr ⟨,x ⟩) or integrate(expr ⟨,x ⟩) returns a primitive of expr with respect to x.
- To evaluate a definite integral, int (or integrate) takes four arguments:
  - expr, an expression.
  - x, the variable.
  - a and b, the bounds of the definite integral.
- int(expr,x,a,b) or integrate(expr,x,a,b) returns the exact value of the definite integral if the computation was successful or an unevaluated integral otherwise.

#### Examples

integrate(x^2)

x3
3

integrate(t^2,t)

t3
3

integrate(x^2,x,1,2)

7
3

integrate(1/(sin(x)+2),x,0,2*pi)

2
3
| π | √ |
3

Int is the inert form of integrate, it prevents evaluation, for example to avoid a symbolic computation that might not be successful if you just want a numeric evaluation, like for example:

evalf(Int(exp(x^2),x,0,1))

or:

evalf(int(exp(x^2),x,0,1))

1.46265174591 |  |  |  |  |  |  |  |  |  |

Let f(x)=x/x2−1+ln(x+1/x−1). Find a primitive of f.

int(x/(x^2-1)+ln((x+1)/(x-1)))

or:

f(x):=x/(x^2-1)+ln((x+1)/(x-1)):; int(f(x))

x ln | ⎛ ⎜ ⎜ ⎝ |
x+1
x−1
| ⎞ ⎟ ⎟ ⎠ | + |
2
2
| ln | ⎪ ⎪ | x2−1 | ⎪ ⎪ | + |
ln | ⎪ ⎪ | x2−1 | ⎪ ⎪
2

Compute ∫2/x6+2x4+x2 dx.

int(2/(x^6+2*x^4+x^2))

2 | ⎛ ⎜ ⎜ ⎜ ⎝ |
−3 x2−2
2 | ⎛ ⎝ | x3+x | ⎞ ⎠
| − |
3
2
| arctanx | ⎞ ⎟ ⎟ ⎟ ⎠

Compute ∫1/sin(x)+sin(2 x) dx.

integrate(1/(sin(x)+sin(2*x)))

2 | ⎛ ⎜ ⎜ ⎝ |
ln | ⎛ ⎜ ⎜ ⎝ |
1−cosx
1+cosx
| ⎞ ⎟ ⎟ ⎠
12
| − |
ln | ⎪ ⎪ ⎪ ⎪ |
1−cosx
1+cosx
| −3 | ⎪ ⎪ ⎪ ⎪
3
| ⎞ ⎟ ⎟ ⎠