FORM GR1268 Question 21

Question:
What is the value of \(\displaystyle\int_{-\pi/4}^{\pi/4}\left(\cos t+\sqrt{1+t^2}\sin^3t\cos^3t\right)dt\)?

(A) \(0\)     (B) \(\sqrt2\)     (C) \(\sqrt2-1\)     (D) \(\displaystyle\frac{\sqrt2}{2}\)     (E) \(\displaystyle\frac{\sqrt2-1}{2}\)

Answer:
(B)

Answer Key:
We should realize that there is no easy antiderivative.

Even if we mess around using \(\sin^2ｔ+\cos^2ｔ=1\) and such,
because of the product of \(\sqrt{1+t^2}\) and trigonometric function,
it is impossible to find antiderivative.

If it is a curve we are familiar with, we may be able to solve geometrically.
However, we have no idea what this curve looks like.

So, let's start with the easy part.
Integrating the first term yields
\[\begin{align}
\int_{-\pi/4}^{\pi/4}\cos t
&= \left.\sin t\right|_{-\pi/4}^{\pi/4}\\
&=\sin\left(\frac{\pi}{4}\right)-\sin\left(-\frac{\pi}{4}\right)\\
&=\frac{1}{\sqrt2}-\left(-\frac{1}{\sqrt2}\right)\\
&=\frac{2}{\sqrt2}\\
&=\sqrt2.
\end{align}\]
If the integral of the second term is zero, the answer is (B).

Is it the case that the integral of the second term is zero?
Fortunately, it is.

Here's why:
\(\sqrt{1+t^2}\) is an even function.
\(\sin^3t\) is an odd function (because cubing just shrinks \(\sin t\)).
\(\cos^3t\) is an even function.
And even \(\times\) odd \(\times\) even \(=\) odd.

The integral of the second term is zero because the function is odd and the limits of the integration is symmetric around zero.