Question

`int_0^(pi/2) root(7)(sin x)/(root(7)(sin x) + root(7)(cos x))`dx = ?

Options

• `pi/8`

• `pi/3`

• `pi/4`

• `pi/2`

Answer 1

`pi/4`

Explanation:

We have,

I = `int_0^(pi/2) root(7)(sin x)/(root(7)(sin x) + root(7)(cos x))`dx

I = `int_0^(pi/2) root(7)(sin (pi/2 - x))/(root(7)(sin (pi/2 - x)) + root(7)(cos (pi/2 - x)))`dx

I = `int_0^(pi/2) (root(7)(cos x))/(root(7)(cos x) + root(7)(sin x))`dx

2I = `int_0^(pi/2) (root(7)(sin x) + root(7)(cos x))/(root(7)(sin x) + root(7)(cos x))` dx

2I = `int_0^(pi/2)` dx

2I = `[x]_0^(pi/2) = pi/2`

I = `pi/4`