1
$\begingroup$

The integral: $$\int_{0}^{\pi/2} \frac{\sin x \cos^5 x}{(1-2\sin^2x\cos^2x)^2}dx$$ has been encountered today while solving a longlish problem at MSE. The question here is: How would one evaluate it?

Addendum

For an interesting use of this integral see my Answer to: Evaluate: $S=\sum_{j=1}^{\infty}\prod_{k=1}^{j}\frac{2k}{j+k+1}$

$\endgroup$
  • 1
    $\begingroup$ Integrating numerically, it seems to be equal to $\frac{\pi}{8}$. $\endgroup$ – Toby Mak Mar 18 at 12:26
3
$\begingroup$

Define $u=2\cos^{2}{(x)}-1$, then

$$ \begin{aligned} \int_{0}^{\frac{\pi}{2}}{\frac{\sin{(x)}\cos^{5}{(x)}}{(1-2\sin^{2}{(x)}\cos^{2}{(x)})^{2}}dx}&=\frac{1}{4}\left(\int_{-1}^{1}{\frac{du}{u^{2}+1}}+\int_{-1}^{1}{\frac{2u\ du}{(u^{2}+1)^{2}}}\right)\\ \\ &=\frac{\pi}{8} \end{aligned} $$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I don't see how the integral in the question and your integral $(\sin x \cos^4 x) \cdots$ are the same. $\endgroup$ – Toby Mak Mar 18 at 13:21
  • 1
    $\begingroup$ @TobyMak whoops missed sth, fixed it $\endgroup$ – Rezha Adrian Tanuharja Mar 18 at 13:24
3
$\begingroup$

$$I=\int_{0}^{\pi/2} \frac{\sin x \cos^5x}{(1-2\sin^2 x \cos^2 x)^2} dx$$ Next using $\int_{0}^{a} f(x) dx= \int_{0}^{a} f(a-x) dx,$ we get $$I=\int_{0}^{\pi/2} \frac{\sin^5 x \cos x}{(1-2\sin^2 x \cos^2 x)^2} dx$$ Adding the two integrals, we get So $$2I=\int_{0}^{\pi/2} \frac{\sin x \cos x(\sin^4 x+ \cos^4 x)}{(1-2\sin^2 x \cos^2 x)^2}=\int_{0}^{\pi/2} \frac{\sin x \cos x}{(1-2\sin^2 x \cos^2 x)}dx=\int_{0}^{\pi/2}\frac{1}{2} \frac{\sin 2x dx}{1+\cos^2 2x}$$ $$ \text{domain halved,}~\implies 2 I=\int_{0}^{\pi/4}\frac{2\sin 2x dx}{1+\cos^2 2x}=-\int_{1}^{0}\frac{dt}{1+t^2}=\frac{\pi}{4} \implies I=\frac{\pi} {8} $$ Lastly we have used $\cos 2x=t.$

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

Let $u= \cos x$, $du = - \sin x \ dx$:

$$\int_{0}^{\pi/2} \frac{\sin x \cos^5 x}{(1-2\sin^2x\cos^2x)^2}dx$$ $$= -\int_{1}^{0} \frac{du \ (u^5)}{(1-2(1-u^2)u^2)^2}$$ $$= \int_{0}^{1} \frac{u^5}{(1-2u^2+2u^4)^2} \ du$$

Then substitute again: let $v = u^2, dv = 2u\ du$:

$$= \frac{1}{2} \int_{0}^{1} \frac{v^2 \ dv}{(1-2v+2v^2)^2}$$

and do (not really) partial fractions:

$$ \frac{v^2 }{(1-2v+2v^2)^2} = \frac{a}{(1-2v+2v^2)} + \frac{b}{(1-2v+2v^2)^2}$$ $$ v^2= a(1-2v+2v^2) + b$$ $$a = \frac{1}{2} \Rightarrow v^2 = \frac{1}{2}-v+v^2+b$$ $$b = v - \frac{1}{2}$$

So we have:

$$\frac{1}{4} \int_{0}^{1} \frac{dv}{(1-2v+2v^2)} + \frac{1}{2} \int_{0}^{1} \frac{v \ dv}{(1-2v+2v^2)^2} - \frac{1}{2} \int_{0}^{1} \frac{\ dv}{(1-2v+2v^2)^2}$$

For the first integral, complete the square which resolves to the standard $\arctan$ integral. For the second integral, let $w = 2v - 1$ (AoPS), and for the third integral, complete the square, then substitute $w = \frac{\tan v}{\sqrt 2}$ where you can use $\tan^2 w + 1 = \sec^2 w$.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Amazing : the tangent half angle substitution works. Let $x=2\tan^{-1}(t)$ to get $$I=-4\int\frac{ t \left(t^2-1\right)^5 \left(t^2+1\right)}{\left(t^8-4 t^6+22 t^4-4 t^2+1\right)^2}\,dt$$ Now, a non-trivial substitution $$t=\sqrt{1+\frac{2 \left(\sqrt{z+1}\right)}{z}}\implies dt=-\frac{1}{2} \sqrt{\frac{z+ 2 \left(\sqrt{z+1}\right)}{z^3(z+1)}}\,dz$$ $$I=\frac{1}{2} \int \frac{dz}{ \left(z^2+1\right)^2}=\frac{1}{4} \left(\frac{z}{z^2+1}+\tan ^{-1}(z)\right)$$ Back to $t$ $$I=\frac{t^2 \left(t^2-1\right)^2}{t^8-4 t^6+22 t^4-4 t^2+1}+\frac{1}{4} \tan ^{-1}\left(\frac{4 t^2}{\left(1-t^2\right)^2}\right)$$

Expanded as a series around $t=1$ gives for the definite integral $$\frac{\pi }{8}-\frac{1}{6} (1-t)^6+O\left((1-t)^7\right)$$

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

Seeing that cosine to an odd power, I would factor out one cosine to use with the derivative, then write the rest of the integrand in terms of sine: $\int_0^{\pi/2}\frac{sin(x)cos^5(x)}{1- 2cos^2(x)sin^2(x)}dx= \int_0^{\pi/2}\frac{sin(x)cos^4(x)}{1- 2cos^2(x)sin^2(x)}(cos(x)dx)= \int_0^{\pi/2} \frac{sin(x)((1- sin^2(x))^2}{1- 2(1- sin^2(x))sin^2(x)}(cos(x)dx)$.

Let u= sin(x) so that du= cos(x)dx. When x= 0, u= 0 and when $x= \pi/2$ u= 1 so the integral becomes $\int_0^1 \frac{u(1- u^2)^2}{1- 2(1- u^2)u^2}du$

| cite | improve this answer | |
$\endgroup$

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.