1
$\begingroup$

Please help me to evaluate definite integral $$\int_{^{-\pi}/_2}^{^\pi/_2}\cos^{2}\left(\theta\right)\,{\rm d}{\theta}$$ Also there was a hint: Use the double angle formula $\cos\left(2x\right)=2\cos^{2}\left(x\right) - 1$.

I know how to do an integral, but I don't know how to evaluate it with this hint. Please help me.

$\endgroup$
  • $\begingroup$ If $\cos 2x = 2\cos^2 x-1$, what is $\cos^2 x$ equal to? Try making this substitution in the integral and see if that helps. $\endgroup$ – rogerl Jul 27 '14 at 18:59
  • $\begingroup$ The hint is to help you rewrite $\cos^2\theta$ in terms of $\cos 2\theta$ $\endgroup$ – MPW Jul 27 '14 at 19:02
  • $\begingroup$ See Wallis' integrals. $\endgroup$ – Lucian Jul 27 '14 at 19:14
  • $\begingroup$ oh, now i remember, 5 years after math lessons, forgot all i've learned. $\endgroup$ – aine Jul 27 '14 at 19:15
  • $\begingroup$ @aine : Maybe you're too quick to "accept" an answer and run away. The question has more interesting answers than the one you "accepted". $\endgroup$ – Michael Hardy Jul 27 '14 at 19:33
4
$\begingroup$

From the hint, $\cos^2 x= \frac{1}{2}(\cos 2x+1)$ so our integral becomes $$\int_{-\pi/2}^{\pi/2} \frac{1}{2}(\cos 2x+1)dx= \left. \left (\frac{1}{4}\sin 2x+\frac{1}{2}x \right ) \right |_{-\pi/2}^{\pi/2}=\pi/2$$

$\endgroup$
  • $\begingroup$ great, thanks for detailed explanation. $\endgroup$ – aine Jul 27 '14 at 19:16
  • $\begingroup$ Not the slickest was to do it. You don't need antiderivatives at all to evaluate this integral. $\endgroup$ – Michael Hardy Jul 27 '14 at 19:32
7
$\begingroup$

Note that $$\int_{- \pi /2}^{\pi / 2} \cos^2 \theta d \theta= \int_{- \pi /2}^{\pi / 2} \sin^2 \theta d \theta \ $$ and that $$\pi = \int_{- \pi /2}^{\pi / 2} 1 d \theta = \int_{- \pi /2}^{\pi / 2} (\cos^2 \theta +\sin^2 \theta) d \theta = 2 \int_{- \pi /2}^{\pi / 2} \cos^2 \theta d \theta$$ hence $$\int_{- \pi /2}^{\pi / 2} \cos^2 \theta d \theta= \frac{\pi}{2}$$

$\endgroup$
3
$\begingroup$

$$ \int_{-\pi/2}^{\pi/2} \cos^2\theta\,d\theta + \int_{-\pi/2}^{\pi/2} \sin^2\theta\,d\theta = \int_{-\pi/2}^{\pi/2} 1\,d\theta=\pi. $$ If you can show the two integrals are equal, then they each have to be $\pi/2$.

But they have to be equal since the graph of $\sin^2$ has the same size and shape as that of $\cos^2$ and the interval from $-\pi/2$ to $\pi/2$ is a full period.

$\endgroup$
  • $\begingroup$ yeah thats nice, thank you very much, i'll know that for future, but i need to use the hint in the task...but i could have some interesting discussion with my teacher ;) $\endgroup$ – aine Jul 27 '14 at 19:42
  • $\begingroup$ I love this. __ $\endgroup$ – Kaj Hansen Jul 27 '14 at 21:46
1
$\begingroup$

Here's yet another way: \begin{align} & \int \cos^2\theta\,d\theta = \int(\cos\theta) \Big(\cos\theta\,d\theta\Big) = \int u\,dv = uv-\int v\,du \\[10pt] = {} & -\cos\theta\sin\theta -\int(\sin\theta)\, \Big( -\sin\theta\, d\theta\Big) \\[10pt] = {} & -\cos\theta\sin\theta + \int \sin^2\theta\,d\theta = -\cos^2\theta + \int(1-\cos^2\theta)\,d\theta \\[10pt] = {} & -\cos\theta\sin\theta + \theta -\int\cos^2\theta\,d\theta. \end{align} So we have shown that $$ \int \cos^2\theta\,d\theta = -\cos\theta\sin\theta + \theta -\int\cos^2\theta\,d\theta. $$ Now add the same thing to both sides: \begin{align} & \phantom{{}+{}}\int \cos^2\theta\,d\theta = -\cos\theta\sin\theta + \theta -\int\cos^2\theta\,d\theta. \\[12pt] & {} + \int \cos^2\theta\,d\theta \phantom{{}= -\cos\theta\sin\theta + \theta }{} + \int\cos^2\theta\,d\theta \\[15pt] & \phantom{+{}}2\int\cos^2\theta\,d\theta = -\cos\theta\sin\theta + \theta + \text{constant} \end{align} and then divide both sides by $2$.

$\endgroup$
  • $\begingroup$ Vaguely reminds me of this thread: math.stackexchange.com/questions/798215/… $\endgroup$ – Kaj Hansen Jul 27 '14 at 21:47
  • $\begingroup$ @KajHansen : I don't think this argument fits into that thread. This argument is straightforward and not more involved than the other ones that establish this result. $\endgroup$ – Michael Hardy Jul 28 '14 at 13:22
1
$\begingroup$

If you look at the plot of $y=\cos^2x$, you will notice that the curve is perfectly symmetric around $y=\frac12$ and has period $\pi$, as confirmed by $\cos^2x=\frac12\cos2x+\frac12$.

For this reason, the average value over a period is $\frac12$, and the area under the curve is $\frac\pi2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.