Question

`int_(-pi/4)^(pi/4) 1/(1 - sinx) "d"x` = ______.

Options

• 1

• `sqrt(2)`

• 2

• 3

Answer 1

`int_(-pi/4)^(pi/4) 1/(1 - sinx) "d"x` = 2.

Explanation:

`int_(-pi/4)^(pi/4) 1/(1 - sinx) "d"x = int_(-pi/4)^(pi/4) (1/(1 - sinx) xx(1 + sinx)/(1 + sin x))"d"x`

= `int_(-pi/4)^(pi/4) (1 + sinx)/(1 - sin^2x) "d"x`

= `int_(-pi/4)^(pi/4) (1 + sin^2x)/(cos^2) "d"x`

= `int_(-pi/4)^(pi/4) (sec^2x + secx tanx) d"x`

= `[tanx + secx]_(-pi/4)^(pi/4)`

= `[tan  pi/4 + sec  pi/4] - [tan(- pi/4) + sec(-pi/4)]`

= `(1 + sqrt(2)) - (-1 + sqrt(2))`

= 2