A calculus problem by Hung Woei Neoh

Calculus Level 3

$\large \int_{-\pi/4}^{\pi/4} \dfrac{x^2 \tan x}{1+ \cos^4 x} dx= \, ?$

Source: My high school friend's college calculus course practice question set.

The answer is 0.

4 solutions

Chew-Seong Cheong
Dec 21, 2016

$\begin{aligned} I & = \int_{-\frac \pi 4}^\frac \pi 4 \frac {x^2 \tan x}{1+ \cos^4 x} dx & \small \color{#3D99F6} \text{Using identity } \int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx \\ & = \frac 12 \int_{-\frac \pi 4}^\frac \pi 4 \left( \frac {x^2 \tan x}{1+ \cos^4 x} + \frac {(-x)^2 \tan (-x)}{1+ \cos^4 (-x)} \right) dx \\ & = \frac 12 \int_{-\frac \pi 4}^\frac \pi 4 \left( \frac {x^2 \tan x}{1+ \cos^4 x} - \frac {x^2 \tan x}{1+ \cos^4 x} \right) dx \\ & = \boxed{0} \end{aligned}$

Note: The corollary of the identity only applies for functions that satisfy certain conditions, specifically those in which the value of the integral exists.

For example, we cannot use the identity to conclude that $\int_{-1}^1 \frac{1}{x} \, dx = 0$ , because that value is undefined (due to what happens about the pole 0).

Calvin Lin Staff - 4 years, 5 months ago

Its easier to say that its an odd function. Thats why the integral over $0 -> \dfrac{\pi}{4}$ is (-1)* $\dfrac{-\pi}{4} -> 0$

Md Zuhair - 3 years, 7 months ago

Hung Woei Neoh
Dec 20, 2016

Let $f(x) = \dfrac{x^2 \tan x}{1+\cos^4 x}$

Notice that $f(-x) = \dfrac{(-x)^2 \tan (-x)}{1+ \cos^4 (-x)} = \dfrac{-x^2 \tan x}{1+\cos^4 x} = -f(x)$

This means that the function above is an odd function. And we know that:

If $f(x)$ is odd, continuous and bounded between $-x$ and $x$ , then $\displaystyle \int_{-x}^x f(x) dx= 0$

Therefore, $\displaystyle \int_{-\pi/4}^{\pi/4} \dfrac{x^2 \tan x}{1+ \cos^4 x} dx= \boxed{0}$

(The solution has been edited to reflect this comment. The condition of "continuous and bouned" was added in.)

Note: The claim about odd functions is not true. You also need to add a condition like boundedness. For example, $\int_{-1}^1 \frac{1}{x} \, dx$ does not exist.

Calvin Lin Staff - 4 years, 5 months ago

@Calvin Lin Sir, may I know why not?? The integral is simply ln(x) from -1 to 1 and also, it converges, so the answer is 0 and hence, technically, the given integral exists....!!

Aaghaz Mahajan - 3 years ago

Remember that you're integrating over a critical point. As such, you need to consider the limit:

$\lim_{ (a, b) \rightarrow ( 0^-, 0^+) } \int_{-1} ^ {a} \frac{1}{x} \, dx + \int_b^1 \frac{1}{x} \, dx$

Does this 2-variable limit exist?

Hint: What happens if $a = -b \rightarrow 0^-$ ? What happens if $a = - 2b \rightarrow 0^-$ ?

Calvin Lin Staff - 3 years ago

@Calvin Lin – @Calvin Lin Ohh!!! I am sorry Sir!! I don't know what happened.......I had forgotten to check whether the integral was discontinuous or not...!!

Aaghaz Mahajan - 3 years ago

Vlad Vasilescu
Dec 21, 2016

The easiest way is to see that tan (x) is an odd function on the given symmetric interval , which implies that the inegral is 0.

Manikandan As
Jun 28, 2020

Odd function

A calculus problem by Hung Woei Neoh

Source: My high school friend's college calculus course practice question set.

The answer is 0.

4 solutions

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