How do you evaluate the definite integral #int (4x)/sqrt(x^2+1) dx# from # [-sqrt3, sqrt3]#?

Question

How do you evaluate the definite integral #int (4x)/sqrt(x^2+1)  dx# from # [-sqrt3, sqrt3]#?

Scarlett Allison · Accepted Answer

#int_(-sqrt(3))^sqrt(3) (4x)/sqrt(x^2+1) dx = 0#

#int_(-sqrt(3))^sqrt(3) (4x)/sqrt(x^2+1) dx = 2int_(-sqrt(3))^sqrt(3) (d(x^2+1))/sqrt(x^2+1) dx = 4 [sqrt(x^2+1)]_(-sqrt(3))^sqrt(3)=4(2-2) = 0#

We might have spared ourselves the trouble of doing the calculations, noting that:

#f(x) = (4x)/sqrt(x^2+1)#

is an odd function, that is:

#f(-x) = -f(x)#

If such a function can be integrated over the interval #(-a,a)# the integral is always null as:

#int_(-a)^a f(x)dx = int_(-a)^0 f(x)dx + int_0^a f(x)dx #

Substitute #t=-x# in the first integral:

#int_(-a)^0 f(x)dx = int_a^0 f(-t)d(-t) = -int_0^a f(-t)d(-t) = int_0^a f(-t)dt = - int_0^a f(t)dt#

So we get:

#int_(-a)^a f(x)dx = int_0^a f(x)dx - int_0^a f(x)dx = 0#

Adam Adkins · Answer

undefined To evaluate the definite integral $\int_{-\sqrt{3}}^{\sqrt{3}} \frac{4x}{\sqrt{x^2+1}} \, dx$, you can use a trigonometric substitution. Let $x = \tan(\theta)$, so $dx = \sec^2(\theta) \, d\theta$.

Substitute $x$ and $dx$ in terms of $\theta$ into the integral, and adjust the limits accordingly.

$$
\int_{-\sqrt{3}}^{\sqrt{3}} \frac{4x}{\sqrt{x^2+1}} \, dx = \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \frac{4\tan(\theta)}{\sqrt{\tan^2(\theta) + 1}} \cdot \sec^2(\theta) \, d\theta
$$

Now, simplify and integrate using trigonometric identities and techniques. After integration, evaluate the antiderivative at the upper and lower limits to find the definite integral value.