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$$\int \sin^2t\cos^2t\,dt$$

Since both exponent are pair and $\ge 2$, according to my understanding I should use one of these equality to solve :

  • $\sin^2t = \dfrac{1-\cos2t}{2}$
  • $\cos^2t = \dfrac{1+\cos2t}{2}$
  • $\sin t\cos t = \dfrac{\sin2t}{2}$

I've tried replacing with each of these tree, but I am unable to solve.

What is wrong with my understanding ? Maybe I need to integrate by part after (I also tried but was unable to solve again) ? Or maybe the equality I've chosen are wrong ?

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  • $\begingroup$ On the right-hand sides you have $\cos(2t)$ instead of $\cos^2(t)$. $\endgroup$
    – Mihail
    Feb 28, 2015 at 13:05
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    $\begingroup$ Use the third equation and then you will have an integral with $\sin^2{2t}$ term then use the first equation to get a linear integral with $\cos{4t}$ term. $\endgroup$
    – Rammus
    Feb 28, 2015 at 13:09
  • $\begingroup$ Please look at my edits to your question. $\sin t\cos t$, rather than $sintcost$, is the right way to typeset that expression. Likewise "$\ge 2$" rather than ">=2". ${}\qquad{}$ $\endgroup$
    – Michael Hardy
    Feb 28, 2015 at 13:26
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    $\begingroup$ @MichaelHardy Thanks for the edit, I won't repeat this mistake again. $\endgroup$
    – student
    Feb 28, 2015 at 13:51

4 Answers 4

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\begin{align} & \int \frac{\sin^2 2t}{4} dt \\[6pt] = {} & \int \dfrac{1 - \cos 4t}{8} dt \\[6pt] = {} & \frac{t}{8} - \frac{\sin 4t}{32} + C \end{align}

Here $C$ is the constant of indefinite integral.

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  • $\begingroup$ Should be $\frac{t}{8}$ on the last line. $\endgroup$
    – Rammus
    Feb 28, 2015 at 13:11
  • $\begingroup$ That's exactly it, was never able to realize that $sin^22t$ could be 1-cos4t wow. $\endgroup$
    – student
    Feb 28, 2015 at 13:11
  • $\begingroup$ Oh yes sorry always did integral involving x as variable , thanks $\endgroup$
    – AAkash
    Feb 28, 2015 at 13:12
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Hint: $$\int \sin^2t\cos^2tdt = \int \sin^2t(1-\sin^2t)dt = \int (\sin^2t - \sin^4t)dt$$

Another perspective: $$\int \sin^2t\cos^2tdt = \int \frac{\sin^2(2t)}{4}dt = \int \dfrac{1 - \cos 4t}{8} dt $$

Try either of these approaches; second one is easier to get the answer with.

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Hint

$$\sin^2(t)\cos^2(t)=\big(\sin(t)\cos(t)\big)^2=\frac 14 \sin^2(2t)=\frac 18\big(1-\cos(4t)\big)$$

I am sure that you can take from here.

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First notice that:

$$\int \sin^2(t) \cos^2(t) dt = \frac{1}{4} \int \sin^2(2t) dt.$$ Now let $ x = 2t$, so $\frac{dx}{2} = dt$. Now you have: $$ \frac{1}{8} \int \sin^2(x) dx.$$

The problem is now reduced to calculate the integral of $\sin^2(x)$. Integrating by parts gives:

\begin{align} \int \sin^2(x) dx &= -\cos(x) \sin(x) + \int \cos^2(x) dx \\ \int \sin^2(x) dx &= -\cos(x) \sin(x) + \int 1 -\sin^2(x) dx \\ \int \sin^2(x) dx &= \frac{1}{2} \left( -\cos(x) \sin(x) + x \right) \end{align} Now you only have to substitute the $x$ and multiply the $\frac{1}{8}$.

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