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The question is to evaluate $$\int_0^{\pi/4} (\cos 2x)^{3/2} \cos x \, dx$$ I tried replacing $x$ by $\pi/4 -x$ and solving but couldn't get the answer.please help me in this regard.thanks.

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  • $\begingroup$ Use double angle formula, then $u$-substitution. $\endgroup$ – anon Aug 20 '17 at 18:02
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    $\begingroup$ @anon: to get what, exactly? Related: mathworld.wolfram.com/LemniscateConstant.html $\endgroup$ – Jack D'Aurizio Aug 20 '17 at 19:01
  • $\begingroup$ @JackD'Aurizio I didn't bother to work it out, so I guess to get what you got. $\endgroup$ – anon Aug 20 '17 at 19:06
  • $\begingroup$ This question required more work than I had expected. In my answer below, I reduce it to an evaluation of the Beta function. $\qquad$ $\endgroup$ – Michael Hardy Aug 20 '17 at 20:44
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Set $\cos 2x = \sin^2\theta$. Then

$$ \cos x = \sqrt{\frac{1+\sin^2\theta}{2}} \qquad \text{and} \qquad \left|\frac{dx}{d\theta}\right| = \frac{\sin\theta}{\sqrt{1+\sin^2\theta}}. $$

So we have

$$ \int_{0}^{\frac{\pi}{4}} \cos^{3/2}(2x) \cos x \, dx = \frac{1}{\sqrt{2}} \int_{0}^{\frac{\pi}{2}} \sin^4\theta \, d\theta = \frac{3\pi}{16\sqrt{2}}. $$

Of course, you may recognize that this is simply a shortened version of previous answers.


Alternative solution. Apply the substitution $x = \theta/2$ and exploit the symmetry to write

$$ I = \frac{1}{4} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\cos\theta)^{3/2} e^{i\theta/2} \, d\theta = \frac{1}{8\sqrt{2}} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (1 + e^{2i\theta})^{3/2} e^{-i\theta} \, d\theta. $$

Now we utilize the substitution $e^{i\theta} = iz$. If $\Gamma$ denotes the lower semicircular arc from $-1$ to $1$, the resulting integral is

$$ I = -\frac{1}{8\sqrt{2}} \int_{\Gamma} \frac{(1 - z^2)^{3/2}}{z^2} \, dz. $$

This integrand has the following antiderivative on the punctuated unit disc $\mathbb{D}\setminus\{0\}$:

$$ \frac{d}{dz} \left[ \left( \frac{1}{z} + \frac{z}{2} \right) \sqrt{1-z^2} + \frac{3}{2}\arcsin z \right] = - \frac{(1 - z^2)^{3/2}}{z^2}. $$

(This expression can be easily obtained by applying integration by parts twice. Of course, this back-of-the-envelope computation is then justified since the resulting function is holomoprhic on $\mathbb{D}\setminus\{0\}$ and has the desired derivative.) So $I$ can be computed as

$$ I = \frac{1}{8\sqrt{2}} \left[ \left( \frac{1}{z} + \frac{z}{2} \right) \sqrt{1-z^2} + \frac{3}{2}\arcsin z \right]_{z=-1}^{z=1} = \frac{3\pi}{16\sqrt{2}}.$$

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$$\int_{0}^{\pi/4}\left(\cos(2x)\right)^{3/2}\,dx \stackrel{x\mapsto\frac{z}{2}}{=} \frac{1}{2}\int_{0}^{\pi/2}\left(\cos z\right)^{3/2}\,dz \stackrel{z\mapsto\frac{\pi}{2}-z}{=} \frac{1}{2}\int_{0}^{\pi/2}\left(\sin z\right)^{3/2}\,dz $$ equals: $$ \frac{1}{2}\int_{0}^{1}u^{3/2}(1-u^2)^{-1/2}\,du \stackrel{u\mapsto v^{1/2}}{=} \frac{1}{4}\int_{0}^{1}v^{1/4}(1-v)^{-1/2}\,dv = \frac{1}{4} B\left(\frac{5}{4},\frac{1}{2}\right) $$ or:

$$ \frac{\Gamma\left(\frac{5}{4}\right)\sqrt{\pi}}{4\,\Gamma\left(\frac{7}{4}\right)}=\frac{\Gamma\left(\frac{1}{4}\right)\sqrt{\pi}}{12\,\Gamma\left(\frac{3}{4}\right)}=\color{blue}{\frac{\Gamma\left(\frac{1}{4}\right)^2}{12\sqrt{2\pi}}} $$ clearly related with the lemniscate constant.
In terms of the complete elliptic integral of the first kind,

$$\begin{eqnarray*}\int_{0}^{\pi/4}\left(\cos(2x)\right)^{3/2}\,dx=\frac{1}{3\sqrt{2}}\int_{0}^{\pi/2}\frac{d\theta}{\sqrt{1-\frac{1}{2}\cos^2\theta}}&=&\color{blue}{\frac{\pi}{6\sqrt{2}}\sum_{n\geq 0}\frac{\binom{2n}{n}^2}{32^n}}\\&=&\color{blue}{\frac{\pi}{6\,\text{AGM}(1,\sqrt{2})}}.\end{eqnarray*}$$

The last form is very well-suited for numerical evaluation, leading to $\color{blue}{I\approx 0.43700959238202}$.
In a similar way $$ \int_{0}^{\pi/4}\left(\cos(2x)\right)^{3/2}\cos(x)\,dx\stackrel{x\mapsto\arcsin(t)}{=}\int_{0}^{1/\sqrt{2}}(1-2t^2)^{3/2}\,dt \stackrel{t\mapsto\frac{u}{\sqrt{2}}}{=}\frac{1}{\sqrt{2}}\int_{0}^{1}(1-u^2)^{3/2}\,du$$ equals (through $u\mapsto\sqrt{v}$) $$ \frac{1}{2\sqrt{2}}\int_{0}^{1}v^{-1/2}(1-v)^{3/2}=\frac{1}{2\sqrt{2}}B\left(\frac{1}{2},\frac{5}{2}\right)=\color{blue}{\frac{3\pi}{16\sqrt{2}}}.$$

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  • $\begingroup$ ok, Maybe it's fun to do the whole problem, but is it not clear that this student's difficulty probably lay in either failure to recognize an occasion to use a double-angle formula or a half-angle formula or the like, or in not knowing what substitution to do? $\qquad$ $\endgroup$ – Michael Hardy Aug 20 '17 at 19:27
  • $\begingroup$ @MichaelHardy: are we focusing on the question or on the difficulties of the problem poser? The question asks for an evaluation. $\endgroup$ – Jack D'Aurizio Aug 20 '17 at 19:35
  • $\begingroup$ This is not the integrand in the OP. $\endgroup$ – B. Goddard Aug 20 '17 at 19:54
  • $\begingroup$ The original integral has a $\cos x$ between the $3/2$ power and the $dx.$ $\endgroup$ – B. Goddard Aug 20 '17 at 20:03
  • $\begingroup$ @B.Goddard: oh, my bad. All right, I am going to tackle the wanted version too. Thanks for pointing that out. $\endgroup$ – Jack D'Aurizio Aug 20 '17 at 20:09
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$$ \cos(2x) = 1 - 2\sin^2 x. $$ \begin{align} \int_0^{\pi/4} (\cos(2x))^{3/2} \cos x\,dx & = \int_0^{\pi/4} \left( 1-2\sin^2 x \right)^{3/2} \overbrace{\Bigg( \cos x\,dx\Bigg)}^{\large\text{This will be } du.} \\[10pt] & = \int_0^{1/\sqrt 2} (1-2u^2)^{3/2} \, du \\[10pt] & = \int_0^1 (1-w^2)^{3/2} \, \frac{dw}{\sqrt 2} \\[10pt] & = \frac 1 {2\sqrt 2} \int_{-1}^1 (1-w^2)^{3/2} \, dw \text{ (since this is an even function)} \\[10pt] & = \frac 1 {2\sqrt 2} \int_{-1}^1 (1+w)^{3/2}(1-w)^{3/2} \, dw \\[10pt] & = \frac 1 {2\sqrt 2} \int_0^1 (2v)^{3/2}(2(1-v))^{3/2} (2 \, dv) \\[10pt] & = \frac 8 {\sqrt 2} \int_0^1 v^{3/2} (1 - v)^{3/2} \, dv \\[10pt] & = \frac 8 {\sqrt 2} \operatorname{B}\left( \frac 5 2, \frac 5 2 \right) = \frac 8 {\sqrt 2} \cdot\frac{\Gamma(5/2)\Gamma(5/2)}{\Gamma(5)} = \frac {3\pi} {16\sqrt 2}. \end{align}

APPENDIX: \begin{align} \int_0^1 x^{\alpha-1}(1-x)^{\beta-1} \, dx & = \operatorname{B}(\alpha,\beta) \quad \text{(the Beta function)} \\[10pt] & = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)} \\[10pt] \text{and } \Gamma(\alpha+1) & = \alpha\Gamma(\alpha), \\[10pt] \text{so that } \Gamma\left( \frac 5 2 \right) & = \frac 3 2 \Gamma\left( \frac 3 2 \right) = \frac 3 2 \cdot \frac 1 2 \cdot \Gamma\left( \frac 1 2 \right) = \frac 3 2 \cdot \frac 1 2 \cdot \sqrt \pi. \end{align}

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  • $\begingroup$ @JackD'Aurizio : I'm not too surprised if I missed a detail. Stand by..... $\endgroup$ – Michael Hardy Aug 20 '17 at 20:21
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    $\begingroup$ @JackD'Aurizio : You're quite right: $$ \int_0^1 x^{\alpha-1}(1-x)^{\beta-1} \, dx = \operatorname{B} (\alpha,\beta) \ne \operatorname{B}(\alpha-1,\beta-1). $$ $\endgroup$ – Michael Hardy Aug 20 '17 at 20:36
  • $\begingroup$ That is Legendre's fault, $\Gamma(t)$ should have been $\int_{0}^{+\infty}x^t e^{-x}\,dx$ and not $\int_{0}^{+\infty}x^{t-1}e^{-x}\,dx$ as he chose :D $\endgroup$ – Jack D'Aurizio Aug 20 '17 at 20:45
  • $\begingroup$ @JackD'Aurizio : I'm not convinced that's what it should have been. Notice that if $$ f_\alpha(x) = \frac 1 {\Gamma(\alpha)} x^{\alpha-1} e^{-x} \text{ for } x\ge 0 \text{ (and 0 for } x<0) $$ then we have the convolution $$ f_\alpha * f_\beta = f_{\alpha+\beta}. $$ And if the density of a probability distribution is $\Big(\text{constant} \cdot x^{\alpha-1} (1-x)^{\beta-1} \Big) \text{ for } 0\le x\le 1$ then the expected value is $\alpha/(\alpha+\beta). \qquad$ $\endgroup$ – Michael Hardy Aug 20 '17 at 20:49
  • $\begingroup$ @JackD'Aurizio : Can you post an answer to the following question, defending the position you took here? math.stackexchange.com/questions/2400650/… $\endgroup$ – Michael Hardy Aug 20 '17 at 23:34
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Hint: $\cos(2x) = \cos^2(x)-\sin^2(x) = 1-2\sin^2(x)=2\cos^2(x)-1$.

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  • $\begingroup$ Thank you.this simplifies the integral to $\int_0^{1/\sqrt 2} (1-2t^2)^{3/2} dt$ but nothing strikes me on how to solve this $\endgroup$ – user471651 Aug 20 '17 at 18:14
  • $\begingroup$ Use integration by parts. I haven't tried, but I think that $(1-2t^2)^{3/2} = (1-2t^2) (1-2t^2)^{1/2}$ should help. $\endgroup$ – Math Lover Aug 20 '17 at 18:17
  • $\begingroup$ It is not clear to me how the Pythagorean theorem leads to an actual evaluation of the given integral. $\endgroup$ – Jack D'Aurizio Aug 20 '17 at 18:56
  • $\begingroup$ @user471651 Integrate this $t$ integral by parts twice, using $dv = dt.$ It might simplify if you substitute $z = \sqrt{2}{t}$ first. $\endgroup$ – B. Goddard Aug 20 '17 at 19:59

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