Trigonometric Definite Integral

Question

$$\int _0^{\frac{\pi }{2}}\sin^2x\cos^2x\,dx$$

I have been trying to solve this integral using the various trig identities but none worked. When I went on one of the online integral calculators it said that "hyperbolic identities" needed to be used. I have never encountered these before. Symbolab said it was an invalid input.

In the solutions manual, this was the first step:

$$\int _0^{\frac{\pi }{2}}\sin^2x\cos^2x\,dx=\int _0^{\frac{\pi }{2}}\frac{1}{4}\left(4\sin^2x\cos^2x\right)\,dx$$

I am already confused as to how this was done. It does not seem to be related to any usual trig identity. Can it be done without these "hyperbolic identities"?

Any help?

Answer 1

It is the very common identity $\sin(2x) = 2\sin(x)\cos(x)$. This leads to following:

$$\frac{1}{4}\int_{0}^{\pi/2} \sin^2(2x) dx = \frac{1}{8} \int_{0}^{\pi/2} (1-\cos(4x)) dx\\ = \frac{\pi}{16}$$

Answer 2

$\begin{align}\sin^2 x\cos^2 x&=\left(\frac{\text{e}^{ix}+\text{e}^{-ix}}{2}\right)^2\left(\frac{\text{e}^{ix}-\text{e}^{-ix}}{2i}\right)^2\\ &=-\frac{\text{e}^{4ix}}{16}-\frac{\text{e}^{-4ix}}{16}+\frac{1}{8}\\ &=-\frac{1}{8}\left(\frac{\text{e}^{i4x}+\text{e}^{-i4x}}{2}\right)+\frac{1}{8}\\ &=-\frac{\cos(4x)}{8}+\frac{1}{8} \end{align}$

Therefore,

$\begin{align}\int_0^{\frac{\pi}{2}}\sin^2 x\cos^2 x\,dx&=\left[\frac{x}{8}-\frac{\sin(4x)}{32}\right]_0^{\frac{\pi}{2}}\\ &=\boxed{\frac{\pi}{16}} \\\end{align}$

Answer 3

There are no "hyperbolic identities" involved, you must have made incorrect input to the calculator.

The "first step" is completely trivial (check again) and just hinting you to use the identity $2\sin x\cos x=\sin2x$.

To solve the integral, you will need to get rid of the square in $\sin^22x$ by a second transform.

Or, by parts, $$I:=\int \sin^2x\,dx=-\cos x\sin x+\int\cos^2x\,dx.$$

Then

$$I+I=-\cos x\sin x+\int(\cos^2x+\sin^2x)\,dx=-\cos x\sin x+x.$$

(Restore the proper factors.)