Find $\int_0^\frac{\pi}{2} \frac {\theta \cos \theta } { \sin \theta + \sin ^ 3 \theta }\:d\theta$

Question

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?

Answer 1

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

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

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

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

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

Answer 2

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

Answer 3

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

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

Answer 4

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.

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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

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

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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.

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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

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

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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.

Answer 5

\begin{align} &\int_0^{\frac{\pi}{2}}\dfrac{x \cos x}{\sin{x}+\sin^3 x}\, dx \\ =& \ \frac12\int_0^{\frac{\pi}{2}}x\ d\left(\ln\frac {2\sin^2x} {1+\sin^2x}\right) \overset{ibp}= -\frac12 \int_0^{\frac{\pi}{2}} \ln\frac {2\sin^2x} {1+\sin^2x}dx\\ = & \ \frac12 \int_0^{\frac{\pi}{2}}\int_0^1 \frac {\sin^2 x} {(1+a\cos^2x)(1-a+2a\cos^2x)}da \ {dx}\\ =&\ \frac\pi4\int_0^1 \frac1{\sqrt{1-a^2}(1+\sqrt{1+a})}\ da=\frac\pi4\ln\left(\frac32+\sqrt2\right) \end{align}

Answer 6

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

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

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

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.