Integrating standard functions
In a nutshell
It is possible to integrate functions other than those of the form axn. Knowing some standard integrals will help solve problems quicker.
Standard integrals
Here is a list of integrals you need to know. In it, n and C are constants.
function
| integral
|
| xn
| ∫xndx=n+11xn+1+C,n=1 |
| x1
| ∫x1dx=ln∣x∣+C
|
| ex
| ∫exdx=ex+C
|
| sin(x)
| ∫sin(x)dx=−cos(x)+C |
| cos(x)
| ∫cos(x)dx=sin(x)+C
|
| sec2(x)
| ∫sec2(x)dx=tan(x)+C
|
| cosec2(x)
| ∫cosec2(x)dx=−cot(x)+C
|
| sec(x)tan(x)
| ∫sec(x)tan(x)dx=sec(x)+C
|
| cosec(x)cot(x)
| ∫cosec(x)cot(x)dx=−cosec(x)+C
|
Note: The trigonometric integrals assume that x is in radians, not degrees.
Example 1
Compute the following integrals:
i) ∫4x1dx
ii) ∫2π23π(sin(x)−4cot(x)−cos(x))dx
Part i):
Use the fact that ∫kf(x)dx=k∫f(x)dx to turn the integral into one of the standard integrals.
∫4x1dx=∫41×x1dx=41∫x1dx=41ln∣x∣+C
∫4x1dx=41ln∣x∣+C
Part ii):
Use the fact that ∫(f(x)+g(x))dx=∫f(x)dx+∫g(x)dx together with the procedure for definite integration.
∫2π23π(sin(x)−4cot(x)−cos(x))dx=∫2π23π(−4cot(x)×sin(x)1−cos(x))dx=∫2π23π(−4cot(x)cosec(x)−cos(x))dx=−4∫2π23πcot(x)cosec(x)dx−∫2π23πcos(x)dx=[−4(−cosec(x))−(sin(x))]2π23π=[4cosec(x)−sin(x)]2π23π=(sin(23π)4−sin(23π))−(sin(2π)4−sin(2π))=(−14−(−1))−(14−(1))=(−3)−(3)=−6
∫2π23π(sin(x)−4cot(x)−cos(x))dx=−6
