What is the definite integral of $\frac{1}{36 + x^{2}}$ $1/(36+x^2)$ with bounds $[0, 6]$ $[0, 6]$ ?

Question

What is the definite integral of $\frac{1}{36 + x^{2}}$ $1/(36+x^2)$ with bounds $[0, 6]$ $[0, 6]$ ?

1 Answer

Impact of this question

6792 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-10T20:59:30

$\frac{π}{24}$ $pi/24$

Explanation:

There is a standard integral: $\int (\frac{1}{a^{2} + x^{2}}) d x = \frac{1}{a} {tan}^{- 1} (\frac{x}{a}) + c$ $int(1/(a^2+x^2))dx = 1/atan^-1(x/a)+c$

So $\int_{0}^{6} (\frac{1}{36 + x^{2}}) d x = {[\frac{1}{6} {tan}^{- 1} (\frac{x}{6})]}_{0}^{6}$ $int_0^6(1/(36+x^2))dx = [1/6tan^-1(x/6)]_0^6$

$= (\frac{1}{6} {tan}^{- 1} (1)) - (\frac{1}{6} {tan}^{- 1} (0)) = \frac{1}{6} \cdot \frac{π}{4} - \frac{1}{6} \cdot 0 = \frac{π}{24}$ $=(1/6tan^-1(1))-(1/6tan^-1(0)) = 1/6*pi/4 - 1/6*0 = pi/24$

Science

Math

Humanities

... and beyond

What is the definite integral of $\frac{1}{36 + x^{2}}$ $1/(36+x^2)$ with bounds $[0, 6]$ $[0, 6]$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

What is the definite integral of 136+x21/(36+x^2) with bounds [0,6][0, 6]?

1 Answer

Explanation:

Related questions

Impact of this question

What is the definite integral of $\frac{1}{36 + x^{2}}$ $1/(36+x^2)$ with bounds $[0, 6]$ $[0, 6]$ ?