Using basic properties of definite integrals, prove that

Using basic properties of definite integrals, prove that $1^{\circ}\ \int_{-1}^1 \arcsin x \, dx=0$; $2^{\circ}\ \int_{-a}^a \sinh x \, dx=0$; $3^{\circ}\ \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \tan x \, dx=0$.

We will use the property that for an odd function $f(x)$ (i.e., $f(-x) = -f(x)$), the integral over a symmetric interval is zero:

\[\int_{-a}^a f(x) \, dx = 0\]

This follows from the fact that:

\[\int_{-a}^a f(x) \, dx = \int_{-a}^0 f(x) \, dx + \int_0^a f(x) \, dx\]

For the first integral, substitute $u = -x$, so $du = -dx$. When $x = -a$, $u = a$; when $x = 0$, $u = 0$:

\[\int_{-a}^0 f(x) \, dx = \int_a^0 f(-u)(-du) = \int_0^a f(-u) \, du = -\int_0^a f(u) \, du\]

Since $f$ is odd, $f(-u) = -f(u)$:

\[\int_{-a}^0 f(x) \, dx = -\int_0^a (-f(u)) \, du = -\int_0^a f(u) \, du\]

Therefore:

\[\int_{-a}^a f(x) \, dx = -\int_0^a f(x) \, dx + \int_0^a f(x) \, dx = 0\]

$1^{\circ}$ Prove $\int_{-1}^1 \arcsin x \, dx = 0$

Check if $f(x) = \arcsin x$ is odd:

We need to verify that $\arcsin(-x) = -\arcsin(x)$.

Let $y = \arcsin(x)$, so $\sin(y) = x$ where $y \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$.

Then $\sin(-y) = -\sin(y) = -x$, which means $-y = \arcsin(-x)$.

Therefore: $\arcsin(-x) = -\arcsin(x)$.

Since $\arcsin x$ is an odd function and the integration interval $[-1, 1]$ is symmetric about zero:

\[\boxed{\int_{-1}^1 \arcsin x \, dx = 0}\]

$2^{\circ}$ Prove $\int_{-a}^a \sinh x \, dx = 0$

Check if $f(x) = \sinh x$ is odd:

Recall that $\sinh x = \frac{e^x - e^{-x}}{2}$.

Then:

\[\sinh(-x) = \frac{e^{-x} - e^x}{2} = \frac{-(e^x - e^{-x})}{2} = -\sinh(x)\]

Since $\sinh x$ is an odd function and the integration interval $[-a, a]$ is symmetric about zero:

\[\boxed{\int_{-a}^a \sinh x \, dx = 0}\]

$3^{\circ}$ Prove $\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \tan x \, dx = 0$

Check if $f(x) = \tan x$ is odd:

Recall that $\tan x = \frac{\sin x}{\cos x}$.

Then:

\[\tan(-x) = \frac{\sin(-x)}{\cos(-x)} = \frac{-\sin x}{\cos x} = -\tan(x)\]

Since $\tan x$ is an odd function and the integration interval $\left[-\frac{\pi}{4}, \frac{\pi}{4}\right]$ is symmetric about zero:

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

Conclusion

All three integrals equal zero because the integrands are odd functions integrated over symmetric intervals about the origin.