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Evaluate the definite integral

$$ I=\int_{0}^{\pi/4}(\cos 2x)^{11/2}\cdot \cos x\;dx $$

My Attempt:

$$ I = \int \left(1-2\sin^2 x\right)^{11/2}\cdot \cos x\;dx $$

Now, substitute $\sin x=t$ with $\cos x \,dx = dt$:

$$ I = \int (1-2t^2)^{11/2}\;dt $$

How can I complete the solution from this point?

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    $\begingroup$ The calculations are very long: But if we apply successively Integration by parts we can reduce the integral to $(1-2t^2)^{\frac{-1}{2}}$ which gives you an $\arcsin(\sqrt 2 t)$ $\endgroup$ – Elaqqad Apr 25 '15 at 13:25
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    $\begingroup$ @Elaqqad. I bet that tedious is just an understatement. $\endgroup$ – Claude Leibovici Apr 25 '15 at 13:26
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Applying the substitution $$s = \sqrt{2} \sin x, \qquad ds = \sqrt{2} \cos x \,dx $$ (which up to a constant is the one you suggest), we get $$\require{cancel} \int_0^{\pi / 4} (\cos 2x)^{11 / 2} \cos x \,dx = \frac{1}{\sqrt{2}} \int_0^1 (1 - s^2)^{11 / 2} ds.$$

Using integration by parts, we can rewrite the integral in $u$ in terms of an integral with a smaller exponent: For a general exponent $\alpha \neq 0$, taking $$u = (1 - s^2)^{\alpha}, \qquad dv = ds$$ gives $$\int \underbrace{(1 - s^2)^{\alpha}}_u \underbrace{ds}_{dv} = \underbrace{(1 - 2s^2)^{\alpha}}_u \underbrace{s}_v - \int \underbrace{s}_v \cdot \underbrace{\alpha (1 - s^2)^{\alpha - 1} \cdot (-2s) \,ds}_{du}.$$ Some (only slightly clever) manipulation of the integral on the r.h.s. gives $$\color{#009f00}{\int (1 - s^2)^{\alpha} ds} = s (1 - s^2)^{\alpha} - 2 \alpha \color{#009f00}{\int (1 - s^2)^{\alpha} ds} + 2 \alpha \int (1 - s^2)^{\alpha - 1} ds ,$$ and solving for our integral gives a reduction formula: $$\color{#009f00}{\int (1 - s^2)^{\alpha} ds} = \frac{1}{2 \alpha + 1} s (1 - s^2)^{\alpha} + \frac{2 \alpha}{2 \alpha + 1} \int (1 - s^2)^{\alpha - 1} ds.$$

If we start with a nonintegral half-integer $\frac{2 m - 1}{2}$, inductively applying this formula $m$ times yields expression for the antiderivative where the only integral expression that occurs is the familiar $$\int (1 - s^2)^{-1/2} ds = \arcsin s + C.$$ In our case, though, we need only the given definite integral, and our expression simplifies in a nice way when we specialize to our limits: $$ \int_0^1 (1 - s^2)^{\alpha} ds = \cancelto{0}{\left.\frac{1}{2 \alpha + 1} s (1 - s^2)^{\alpha}\right\vert_0^1} + \frac{2 \alpha}{2 \alpha + 1} \int_0^1 (1 - s^2)^{\alpha - 1} ds , $$ or a little more readably, $$ \phantom{(\ast)} \qquad \int_0^1 (1 - s^2)^{\alpha} ds = \frac{2 \alpha}{2 \alpha + 1} \int_0^1 (1 - s^2)^{\alpha - 1} ds. \qquad (\ast)$$

Taking $\alpha = \frac{11}{2}$ in $(\ast)$ gives $$\int_0^1 (1 - s^2)^{11 / 2} dt = \frac{11}{12} \int_0^1 (1 - s^2)^{9 / 2} ds,$$ and the integral on the r.h.s. is just the integral on the l.h.s. of the reduction formula with $\alpha = \frac{9}{2}$. Proceeding inductively thus gives $$\int_0^1 (1 - s^2)^{11 / 2} dt = \frac{11}{12} \cdot \frac{9}{10} \cdot \frac{7}{8} \cdot \frac{5}{6} \cdot \frac{3}{4} \cdot \frac{1}{2} \cdot \int_0^1 (1 - s^2)^{-1 / 2} ds.$$ The integral on the r.h.s. is $$\int_0^1 (1 - s^2)^{-1 / 2} ds = \left.\arcsin s \right\vert_0^1 = \frac{\pi}{2} .$$ (Alternatively, stopping one step earlier gives the integral $\int_0^{1} \sqrt{1 - s^2} \,ds$, but this is just one-fourth the area of a unit circle, or $\frac{\pi}{4}$.) Now, putting everything together (and remembering the factor of $\frac{1}{\sqrt{2}}$ introduced by a change of variable to $s$) gives $$\color{#bf0000}{\boxed{\int_0^{\pi / 4} (\cos 2x)^{11 / 2} \cos x \,dx = \frac{1}{\sqrt{2}} \cdot \frac{11}{12} \cdot \frac{9}{10} \cdot \frac{7}{8} \cdot \frac{5}{6} \cdot \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2} = \frac{231 \pi}{2048 \sqrt{2}}}}.$$

Remark An induction argument along the same lines gives the general result $$\int_0^1 (1 - s^2)^{(2m - 1) / 2} ds = \frac{1}{4^m}{{2m}\choose{m}} \cdot \frac{\pi}{2} .$$

Interestingly, $$\frac{1}{4^m}{{2m}\choose{m}}$$ is both (1) the coefficient of $r^m$ in the Maclaurin series for $\frac{1}{\sqrt{1 - r^2}}$ and (2) the probability of getting heads exactly half of the time when flipping a coin $2 m$ times, either of which may well hint toward a slicker way of handling this family of integrals.

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    $\begingroup$ Nice answer, and nice LaTeX lesson ;-) $\endgroup$ – Jean-Claude Arbaut Apr 25 '15 at 14:40
  • $\begingroup$ This is real nice ! $\endgroup$ – Claude Leibovici Apr 25 '15 at 14:44
  • $\begingroup$ Thanks, Claude and Jean-Claude! $\endgroup$ – Travis Apr 25 '15 at 14:45
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Use successive integrations by parts. Here is the first:

$$\int (1-2t^2)^{11/2} dt=t(1-2t^2)^{11/2}-\int \frac{11}{2}t (-4t)(1-2t^2)^{9/2}dt$$ $$=t(1-2t^2)^{11/2}-11\int (1-2t^2-1)(1-2t^2)^{9/2}dt$$ $$=t(1-2t^2)^{11/2}-11\int (1-2t^2)^{11/2}dt+11\int (1-2t^2)^{9/2}dt$$

Hence

$$12\int (1-2t^2)^{11/2} dt=t(1-2t^2)^{11/2}+11\int (1-2t^2)^{9/2}dt$$

When you have lowered the exponent enough, it's easy.


The following integrations yield

$$\int (1-2t^2)^{11/2} dt=\frac{1}{12}t(1-2t^2)^{11/2}+\frac{11}{12}\int (1-2t^2)^{9/2}dt$$

$$\int (1-2t^2)^{9/2} dt=\frac{1}{10}t(1-2t^2)^{9/2}+\frac{9}{10}\int (1-2t^2)^{7/2}dt$$

$$\int (1-2t^2)^{7/2} dt=\frac{1}{8}t(1-2t^2)^{7/2}+\frac{7}{8}\int (1-2t^2)^{5/2}dt$$

$$\int (1-2t^2)^{5/2} dt=\frac{1}{6}t(1-2t^2)^{5/2}+\frac{5}{6}\int (1-2t^2)^{3/2}dt$$

$$\int (1-2t^2)^{3/2} dt=\frac{1}{4}t(1-2t^2)^{3/2}+\frac{3}{4}\int (1-2t^2)^{1/2}dt$$

$$\int (1-2t^2)^{1/2} dt=\frac{1}{2}t(1-2t^2)^{1/2}+\frac{1}{2}\int (1-2t^2)^{-1/2}dt$$

The last is

$$\int \frac{1}{\sqrt{1-2t^2}}dt=\frac{1}{\sqrt{2}}\arcsin (\sqrt2 t)+C$$

Finally, modulo typing mistakes

$$\int (1-2t^2)^{11/2} dt= \frac{1}{12}t(1-2t^2)^{11/2} +\frac{11\cdot1}{12\cdot10}t(1-2t^2)^{9/2} +\frac{11\cdot9\cdot1}{12\cdot10\cdot8}t(1-2t^2)^{7/2} +\frac{11\cdot9\cdot7\cdot1}{12\cdot10\cdot8\cdot6}t(1-2t^2)^{5/2} +\frac{11\cdot9\cdot7\cdot5\cdot1}{12\cdot10\cdot8\cdot6\cdot4}t(1-2t^2)^{3/2} +\frac{11\cdot9\cdot7\cdot5\cdot3\cdot1}{12\cdot10\cdot8\cdot6\cdot4\cdot2}t(1-2t^2)^{1/2} +\frac{11\cdot9\cdot7\cdot5\cdot3\cdot1}{12\cdot10\cdot8\cdot6\cdot4\cdot2}\frac{1}{\sqrt{2}}\arcsin (\sqrt2 t)+C $$

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    $\begingroup$ The only requirement is to be patient, I guess ! Cheers :-) $\endgroup$ – Claude Leibovici Apr 25 '15 at 13:36
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    $\begingroup$ I might suggest OP repeat this calculation with general exponent, that is, product a reduction formula that gives $\int (1 - 2 t^2)^{(2m + 1) / 2} dt$ in terms of $\int (1 - 2 t^2)^{(2m - 1) / 2} dt$. Then, one needs only carry out this one i.b.p. computation, apply the result formula five times, which is mostly a matter of arithmetic, and handle the last integral, namely, $\int (1 - 2t^2)^{1 / 2} dt$ manually. $\endgroup$ – Travis Apr 25 '15 at 13:36
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    $\begingroup$ @Travis. As you suggested, I checked : in the book by I.S. Gradshteyn and I.M. Ryzhik, there is a reduction formula for $$\int \sqrt{(a+bx+cx^2)^{2n+1}}dx$$ (see on page 94) $\endgroup$ – Claude Leibovici Apr 25 '15 at 14:00
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    $\begingroup$ @ClaudeLeibovici Thanks for looking this up. I've written up an answer of my own that include the reduction formula for the particular $a, b, c$ that occur here (though of course this essentially Jean-Claude's computation above with constants replaced by appropriate expressions in $n$). This, together with the observation that the "non-integral" term of the r.h.s. of the i.b.p. formula is always zero for this particular definite integral makes doing the original integral less that totally unpleasant. $\endgroup$ – Travis Apr 25 '15 at 14:34
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I would suggest a different change of variables -

Let $t=\cos(2x)$, so $dt=-4\cos(x)\sin(x)dx$ and we have $I=\int_0^{1} t^{11/2}\cdot (dt/4\sin(x))dt$

since $t=\cos(2x)=1-2\sin(x)^2$ we have $\sin(x)=\sqrt{(1-t)/2}$, and the integral reduces to:

$I=(\sqrt2/4)\int_0^{1} t^{11/2}(1-t)^{-1/2}dt$ which in my opinion is easier with integration by parts, comparing to the integral you already have.

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  • $\begingroup$ The last integral in the answer can be expressed in terms of the Beta function and, since the exponents are half-integral, in terms of special values of the Gamma function. $\endgroup$ – Travis Oct 28 '18 at 11:29
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Here's an efficient method using special functions.

Applying the substitution $$t = 2 \sin^2 x, \qquad ds = 4 \sin x \cos x \,dx $$ gives $$\int_0^{\pi / 4} (\cos 2x)^{(2 m - 1) / 2} \cos x \,dx = \frac{1}{2 \sqrt{2}} \int_0^1 t^{-1/2} (1 - t)^{(2 m - 1) / 2} dt .$$ By the definition of and then a standard identity for the Beta function $\textrm{B}$, $$\int_0^1 t^{-1/2} (1 - t)^{(2 m - 1) / 2} dt = \textrm{B} \left(\tfrac{1}{2}, m + \tfrac{1}{2}\right) = \frac{\Gamma(\tfrac{1}{2}) \Gamma(m + \frac{1}{2})}{\Gamma(m + 1)} .$$ where $\Gamma$ is the Gamma function, but we can easily handle all three of the factors $\Gamma(\,\cdot\,)$:

  • By definition, $\Gamma\left(\frac{1}{2}\right) = \int_0^{\infty} q^{-1/2} e^{-q} \,dq ,$ and the substitution $q = r^2, dq = 2 r \, dr$ gives (by symmetry, one-half of) a Gaussian integral, yielding $$\Gamma\left(\tfrac{1}{2}\right) = \sqrt{\pi} .$$
  • Applying the identity $\Gamma(z) = z \Gamma(z - 1)$ a total of $m$ times gives $$\Gamma\left(m + \tfrac{1}{2}\right) = \frac{(2 m)!}{4^m m!} \Gamma\left(\tfrac{1}{2}\right) = \frac{(2 m)!}{4^m m!} \sqrt\pi .$$
  • Since $m$ is an integer, $\Gamma(m + 1) = m !$.

Assembling these pieces (and remembering the factor of $\frac{1}{2 \sqrt{2}}$ in the second display equation) gives $$\int_0^{\pi / 4} (\cos 2x)^{(2 m - 1) / 2} \cos x \,dx = \frac{1}{2 \sqrt{2}} \frac{\left(\sqrt\pi\right)\left(\tfrac{(2 m)!}{4^m m!} \sqrt\pi\right)}{m!} = \frac{\pi}{2 \sqrt{2}} \cdot \frac{1}{4^m} {{2 m} \choose m} .$$

In our case, $\frac{2 m - 1}{2} = \frac{11}{2}$, so $m = 6$, and $$\color{#bf0000}{\boxed{\int_0^{\pi / 4} (\cos 2x)^{11 / 2} \cos x \,dx = \frac{\pi}{2 \sqrt{2}} \cdot \frac{1}{4^6} {{12} \choose 6} = \frac{231 \pi}{2048 \sqrt{2}}}}.$$

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