8
$\begingroup$

My Calc 2 teacher wasn't able to solve this: $$\int_0^\frac{\pi}{2} \frac {\theta \cos \theta } { \sin \theta + \sin ^ 3 \theta }\:d\theta$$

Can someone help me solve this?

$\endgroup$
  • $\begingroup$ This is an old post, but in case anyone sees this, I would recommend getting rid of the $\theta$ in the numerator by differentiating $I(k) = \int_0^\frac{\pi}{2} \frac {\sin(k\theta) } { \sin \theta + \sin ^ 3 \theta }\:d\theta$ and then note that $I(k) = \frac{1}{4}\int_0^{2\pi} \frac {\sin(k\theta) } { \sin \theta + \sin ^ 3 \theta }\:d\theta$ so we get a fairly simple integral by residue theorem by letting $e^{i\theta} = z$. $\endgroup$ – Brevan Ellefsen Jun 8 '18 at 4:57
10
$\begingroup$

First Start with Basic Integration by Parts: $$ \int_0^{\frac{\pi}{2}}\dfrac{x \cos x}{\sin{x}+\sin^3 x}\, dx = -\frac{\pi}{4}\log 2 - \int_0^{\frac{\pi}{2}} \left( \log{\sin{x}} - \frac{\log(1+\sin^2 x)}{2}\, dx \right) $$

$$$$

Since, $$\int_0^{\frac{\pi}{2}} \log\sin x\, dx = -\frac{\pi}{2} \log 2 $$

$$\therefore\int_0^{\frac{\pi}{2}}\frac{x \cos x}{\sin x+\sin^3 x} \, dx = \frac{\pi}{4} \log 2 + \int_0^{\frac{\pi}{2}} \frac{\log(1+\sin^2 x)}{2}\, dx \tag 1$$

$$$$

Now, let (for some $k$): $$I(k) = \int_0^{\frac{\pi}{2}} \frac{\log(1+\sin^2 x + k \sin^2 x)}{2}\, dx \tag 2$$

In order to find $I(k)$:

$$\frac{\partial I}{\partial k} = \int_0^{\frac{\pi}{2}} \frac{\sin^{2}{x}\, dx}{1+\sin^2 x +k\, \sin^2 x} $$

$$\implies I'(k) = \int_0^{\frac{\pi}{2}} \frac{1}{k+1}\left(1-\frac{\sec^2 x}{1+(k+2)\tan^2 x}\right)\, \, dx $$

$$=\frac{x}{k+1}-\frac{\tan^{-1}{(\sqrt{k+2}\,\tan x)}}{(k+1)\, \sqrt{k+2}}\bigg|_0^{\frac{\pi}{2}} = \frac{\pi}{2k+2}\left(1-\frac{1}{\sqrt{k+2}}\right)$$

Now just integrate (using Calculus 1 knowledge): $$I(k)= \frac{\pi}{2}\left(\log{(k+1)}-\log\left(\frac{\sqrt{k+2}-1}{\sqrt{k+2}+1}\right)\right)+C=\frac{\pi}{2}\, \log\left(2\, (\sqrt{k+2}+1)^2\right)+C$$

$$\therefore I(k) = \frac{\pi}{2}\, \log{\left(2 (\sqrt{k+2}+1)^2\right)}+C \tag 3$$

From this step forward, we will solve for $C$:

To find $C$, substitute $k=-1$.

From $(2)$, we have: $$\therefore I(-1) = 0$$

From $(3)$, we have: $$I(-1) = \frac{3\,\pi}{2}\log 2+C$$ $$\implies C= - \frac{3\,\pi}{2}\log 2$$

$$\therefore I(0) = \int_0^{\frac{\pi}{2}} \log(1+\sin^{2}{x})\, dx = \frac{\pi}{2}\, \log\left(2\, (\sqrt{2}+1)^2\right) - \frac{3\,\pi}{2} \log 2$$

Now use $(1)$: $$\therefore\int_0^{\frac{\pi}{2}}\frac{x \cos x\, dx}{\sin x +\sin^3 x} = \frac{\pi}{4}\,\log 2 + \frac{\pi}{4}\, \log\left(2\, (\sqrt{2}+1)^2\right) - \frac{3\,\pi}{4}\log 2$$

$$= \frac{\pi}{4}\log\left(\frac{3+2\,\sqrt{2}}{2}\right)$$

$\endgroup$
  • $\begingroup$ @MichaelHardy Thank You. $\endgroup$ – Yagna Patel Aug 17 '15 at 17:19
  • $\begingroup$ Well done! A big +1 $\endgroup$ – Mark Viola Aug 17 '15 at 17:19
  • 3
    $\begingroup$ +1) I think this answer can also serve as a sort of life-lesson for the high school student: if your calculus teacher is having trouble solving an integral, try differentating under the integral sign. Why does this have such a good chance of working? Because for some some inscrutable reason, THEY DON'T censored TEACH IT! and it's likely the one trick they haven't tried. :< $\endgroup$ – David H Aug 17 '15 at 17:19
  • $\begingroup$ @DavidH That is very true (coming from personal experience). $\endgroup$ – Yagna Patel Aug 17 '15 at 17:21
  • $\begingroup$ @YagnaPatel : I did some edits including the proper use of \tag{}. I also did some clutter reduction on things like \log{(1+{\sin^{3}{x}})} where \log(1+\sin^3 x) is sufficient. ${}\qquad{}$ $\endgroup$ – Michael Hardy Aug 17 '15 at 17:25
4
$\begingroup$

HINT:

$$\int x\dfrac{\cos x}{\sin x(1+\sin^2x)}dx$$

$$=x\int\dfrac{\cos x}{\sin x(1+\sin^2x)}dx-\int\left(\dfrac{dx}{dx}\cdot\int\dfrac{\cos x}{\sin x(1+\sin^2x)}dx\right)dx$$

Set $\sin x=u$

For $\dfrac1{u(u^2+1)}=\dfrac{u^2+1-u^2}{u(u^2+1)}=?$

$\endgroup$
  • 1
    $\begingroup$ I think this is how the exercise was supposed to be done by its authors, to practice on integration techniques. In this respect, it may be worth it to mention that the first step is an integration by parts, with $u=x,\ dv=[\cos x/(\sin x + \sin^3 x)]dx$... Probably the best answer: it highlights the ingretients, and it suggests the way to use them, without solving the whole problem. $\endgroup$ – bartgol Aug 17 '15 at 17:27
1
$\begingroup$

You can transform the integral by parts to get

$$\int_0^{\pi/2} d\theta \frac{\theta \cos{\theta}}{\sin{\theta} (1+\sin^2{\theta})} = \int_0^{\pi/2} d(\sin{\theta}) \left (\frac1{\sin{\theta}}-\frac{\sin{\theta}}{1+\sin^2{\theta}} \right )\theta \\ = -\frac{\pi}{4} \log{2} - \int_0^{\pi/2} d\theta \left [\log{(\sin{\theta})} - \frac12 \log{(1+\sin^2{\theta})} \right ]$$

Now,

$$\int_0^{\pi/2} d\theta \log{(\sin{\theta})} = - \frac{\pi}{2} \log{2}$$

$$\int_0^{\pi/2} d\theta \,\log{(1+\sin^2{\theta})} = \frac{\pi}{2} \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k 2^{2 k}} \binom{2 k}{k} $$

Defining

$$f(x) = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k 2^{2 k}} \binom{2 k}{k} x^k$$

we find that

$$f'(x) = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{2^{2 k}} \binom{2 k}{k} x^{k-1} = \frac1{x} \left (\frac1{\sqrt{1+x}}-1 \right )$$

so that

$$f(x) = \log{\left [\frac{\sqrt{1+x}-1}{ x \left (\sqrt{1+x}+1\right )} \right ]}+C$$

$$\lim_{x \to 0} f(x) = 0 \implies C=\log{4} $$

Thus, putting this all together, we get

$$\int_0^{\pi/2} d\theta \frac{\cos{\theta}}{\sin{\theta} (1+\sin^2{\theta})} = \frac{\pi}{2} \log{(1+\sqrt{2})} - \frac{\pi}{4} \log{2} = \frac{\pi}{4} \log{\left (\frac{3}{2} + \sqrt{2} \right )}$$

$\endgroup$
  • $\begingroup$ (+1) That's what I got as well. $\endgroup$ – Yagna Patel Aug 17 '15 at 17:20
1
$\begingroup$

In the spirit of both (i) the solid answer posted by @yagnapatel and (ii) THIS ANSWER, we proceed here by using the technique of Differentiating Under the Integral Sign to evaluate the integral of interest.


STEP 1:

Let the integral of interest $I$ be given by

$$I=\int_0^{\pi/2} \frac{\theta \cos \theta}{\sin \theta +\sin^3 \theta}\,d\theta \tag 1$$

First, integrating $(1)$ by parts with $u=\theta$ and $v=\log (\sin \theta)-\frac12 \log (1+\sin^2 \theta)$ yields

$$\begin{align} I&=-\frac{\pi}{4}\log 2+\int_0^{\pi/2}\left(\frac12 \log (1+\sin^2 \theta)-\log (\sin \theta)\right)\,d\theta\\\\ &=\frac{\pi}{4}\log 2+\frac12\int_0^{\pi/2} \log (1+\sin^2 \theta)\,d\theta \tag 2 \end{align}$$


STEP 2:

Second, we examine the integral $J(a)$ that is defined as

$$J(a)=\int_0^{\pi/2}\log (a+\sin^2 \theta)\,d\theta$$

and note that $J(1)$ is the integral that appears on the right-hand side of $(2)$.

In addition, we observe that $J(0)=-\pi\log 2$, and $J'(a)$ is given by

$$J'(a)=\int_0^{\pi/2} \frac{1}{a+\sin^2\theta}\,d\theta$$

Now, in THIS ANSWER, I showed that $J'(a)$ is given by

$$\begin{align} J'(a)&=\int_0^{\pi/2}\frac{1}{a+\sin^2\theta}\,d\theta\\\\ &=\text{sgn}(a)\frac{\pi}{2\sqrt{a(a+1)}} \tag 3 \end{align}$$

Note that this result could also be obtained using the well-known Weierstrass Substitution.


STEP 3:

Third, we integrate $(3)$ and make use of $J(0)=-\pi \log 2$ to find that for $a>0$

$$J(a)=\pi \log(\sqrt{a}+\sqrt{1+a})-\pi \log 2$$

and therefore

$$J(1)=\pi\log(1+\sqrt{2})-\pi \log 2 \tag 4$$


STEP 4:

Finally, we substitute $(4)$ into $(2)$ to reveal

$$\bbox[5px,border:2px solid #C0A000]{I=\frac{\pi}{2}\log\left(1+\frac{\sqrt{2}}{2}\right)}$$

which agrees with the result reported by others since $\left(1+\frac{\sqrt{2}}{2}\right)^2=\frac32 +\sqrt{2}$! And we're done.

$\endgroup$
  • $\begingroup$ While both solutions use similar approaches, your path is much cleaner and shorter than the one I took. The use of trig does smooth things out quite a bit. Very nice. (+1) To be honest, I actually prefer this approach over mine. $\endgroup$ – Yagna Patel Aug 17 '15 at 23:46
  • $\begingroup$ @yagnapatel Wow! Thank you very much. As you might know, I had my 4th surgery in the past 15 months 7 weeks ago and while recovering have had yet another complication. Your kind words mean a lot to me. $\endgroup$ – Mark Viola Aug 18 '15 at 1:17
  • $\begingroup$ Please let me know how I can improve my answer. I really want to give you the best answer I can. $\endgroup$ – Mark Viola Aug 28 '15 at 19:37
0
$\begingroup$

$$ \int \frac {\theta \cos \theta } { \sin \theta + \sin ^ 3 \theta} \,d\theta = \underbrace{\int \theta \, dx = \theta x - \int x\,d\theta}_{\text{integration by parts}} $$ $$ dx = \frac { \cos \theta } { \sin \theta + \sin ^ 3 \theta} \,d\theta = \frac{du}{u+u^3} = \frac{du}{u(1+u^2)} = \left( \frac A u + \frac{Bu+C}{1+u^2} \right)\,du $$

$$ \frac{Bu}{1+u^2} \, du = \frac B 2\cdot \frac{dw} w $$

$$ \int \frac C {1+u^2} \, du = C \arctan u +\text{constant} $$

You need to do a bit of algebra to find the three coefficients $A,B,C$, and then put everything together.

I'm getting $A=1$, $B=-1$, $C=0$, so $$ \int \left( \frac 1 u - \frac u {1+u^2}\right)\, du = \log |u| - \frac 1 2 \log(1+u^2) + \text{constant} $$ but we don't need the constant: in integration by parts we need only one antiderivative, not all of them.

This is $$ \log|\sin\theta| - \frac 1 2 \log(1+\sin^2\theta). $$

So the integral is $$ \theta\left( \log|\sin\theta| - \frac 1 2 \log(1+\sin^2\theta) \right) - \int \left( \log|\sin\theta| - \frac 1 2 \log(1+\sin^2\theta) \right) \, d\theta + \text{constant} $$

I haven't taken it beyond that yet, and I don't know whether it can be done is closed form.

However, sometimes it is possible to find a definite integral even if you can't find the indefinite integral.

$\endgroup$
  • $\begingroup$ I think it does have a closed form. Can you please verify my solution? Because I was surprised to see that it came out cleaner than I expected. $\endgroup$ – Yagna Patel Aug 17 '15 at 17:12

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.