What is the value of `int_(-pi//4)^(pi//4) (sin x- tan x)dx`

Options

A. `-1/sqrt2 +ln (1/sqrt2)`
B. `1/sqrt2`
C. `0`
D. `sqrt2`

Answer

To solve the integral \( I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} (\sin x - \tan x) \, dx \), we can use the property of definite integrals that states:

\[ \int_a^b f(x) \, dx = \int_a^b f(a + b - x) \, dx \]

In our case, \( a = -\frac{\pi}{4} \) and \( b = \frac{\pi}{4} \). Therefore, \( a + b = 0 \), and we can rewrite the integral as:

\[ I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \left( \sin x - \tan x \right) \, dx = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \left( \sin(0 - x) - \tan(0 - x) \right) \, dx \]

This simplifies to:

\[ I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \left( -\sin x + \tan x \right) \, dx \]

Now, we can express this as:

\[ I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} (-\sin x + \tan x) \, dx \]

Now, we can combine the two integrals:

\[ 2I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} (\sin x - \tan x) \, dx + \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} (-\sin x + \tan x) \, dx \]

This results in:

\[ 2I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \left( (\sin x - \tan x) + (-\sin x + \tan x) \right) \, dx \]

The terms inside the integral simplify to zero:

\[ 2I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} 0 \, dx = 0 \]

Thus, we find:

\[ 2I = 0 \implies I = 0 \]

Therefore, the value of the integral is:

\[ \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} (\sin x - \tan x) \, dx = 0 \]