int_0^1 1/(x^2+1) = ?

2 Answers
Dec 14, 2017

pi/4

Explanation:

First thing to sonsider is:

int 1/(x^2+1) dx

Make a substitution:

x = tantheta

=> dx = sec^2 theta d theta

=> int 1 / (tan^2 theta +1 ) sec^ 2 d theta

We must use the identity:

1 + tan^2 x = sec^2 x

=> int sec^2 theta / ( sec^2 theta ) d theta

=> int d theta

=> theta + c

Use: tan^(-1) x = theta

= tan^(-1) x + c

Now considering limits from 0 to 1

=> tan^(-1) (1) - tan^(-1)( 0)

=> pi/4 - 0

=> pi/4

Dec 14, 2017

pi/4

Explanation:

int_0^1 1/(x^2+1)dx

we do this by substitution

x=tanu

=>dx=sec^2udu

change of limits

x=0=>u=tan^(-1)0=0

x=1=>u=tan^(-1)1=pi/4

the integral becomes

int_0^(pi/4)1/cancel(tan^2u+1)xxcancel(sec^2u)du

=int_0^(pi/4)du=[u]_0^(pi/4)

=pi/4=0=pi/4