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This question already has an answer here:

I tried the trigonometric substitution $y^2 = \tan \theta, sec^2\theta = 1 + y^4$

But now I'm stuck with $\frac12 \int \frac{\sqrt{\sin \theta}}{(\cos\theta)^{\frac92} } d \theta$

I ran out of imagination as what to try now

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marked as duplicate by Norbert, user17762, Henry T. Horton, Start wearing purple, Stahl May 14 '13 at 21:03

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    $\begingroup$ You can always find a primitive by partial fractions: $1+y^4=(1+y\sqrt{2}+y^2)(1-\sqrt{2}y+y^2)$. Complicated calculations, but in the end it's some arctangent. $\endgroup$ – egreg May 14 '13 at 20:09
  • $\begingroup$ try partial fraction $\endgroup$ – iostream007 May 14 '13 at 20:10
  • $\begingroup$ For @egreg hint you can see here $\endgroup$ – Cortizol May 14 '13 at 20:10
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    $\begingroup$ Btw. (just saying) complex method is useful here $\endgroup$ – Cortizol May 14 '13 at 20:14
  • $\begingroup$ I believe you have made a replacement error. Your substitutions are correct, but you should have $2y \ dy = \sec^2 \theta \ d\theta $ or $dy = \frac{\sec^2 \theta \ d\theta}{2y}$ . This makes your integral $\int \frac{\sec^2 \theta \ d\theta}{2y \ \sec^2 \theta}$ or $\frac{1}{2} \int \frac{d\theta}{\sqrt{\tan \theta}} $ , which is manageable, but not pretty (that is, about as unpleasant as the partial fraction decomposition)... $\endgroup$ – colormegone May 14 '13 at 20:35
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Note that $$\int_1^{\infty} \dfrac{dy}{1+y^4} = \int_0^1 \dfrac{y^2dy}{1+y^4}$$ Hence, $$I=\int_0^{\infty} \dfrac{dy}{1+y^4} = \int_0^1 \dfrac{1+y^2}{1+y^4}dy$$ We have $y^4+1 = (y^2+1+y\sqrt2)(y^2+1-y\sqrt2)$. Hence, $$1+y^2= \dfrac{(y^2+1+y\sqrt2) + (y^2+1-y\sqrt2)}2$$ Hence, we get that \begin{align} I & = \dfrac12\int_0^1\dfrac{dy}{1+y^2-y\sqrt2} + \dfrac12\int_0^1\dfrac{dy}{1+y^2+y\sqrt2}\\ & = \dfrac12\int_0^1\dfrac{dy}{\left(y-\dfrac1{\sqrt2}\right)^2+\left(\dfrac1{\sqrt2} \right)^2} + \dfrac12\int_0^1\dfrac{dy}{\left(y+\dfrac1{\sqrt2}\right)^2+\left(\dfrac1{\sqrt2} \right)^2}\\ & = \dfrac1{\sqrt2} \left(\arctan \left(y\sqrt2-1\right) + \arctan \left(y\sqrt2+1\right) \right)_{y=0}^1\\ & = \dfrac1{\sqrt2} \left(\arctan(\sqrt2 - 1)+\arctan(\sqrt2+1)\right) = \dfrac1{\sqrt2} \left(\arctan\left(\dfrac1{1+\sqrt2} \right)+\arctan(1+\sqrt2)\right)\\ & = \dfrac1{\sqrt2} \times \dfrac{\pi}2 = \dfrac{\pi}{2 \sqrt2} \end{align}

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$$\int_0^\infty \frac{dx}{x^4+1} = \frac{1}{2}\int_{-\infty}^\infty \frac{dx}{x^4+1}$$

The latter integral is trivial by means of contour integration.

Let $C$ be the canonical semicircle contour, along the real axis from $-R$ to $-R$ and around the semicircle $Re^{i\theta}$ for $\theta \in [0,\pi]$. Letting $R \to \infty$, we consider the function on the arc ($|z| = R$):

$$\left|\frac{1}{x^4+1}\right|\le \frac{1}{|x^4+1|} \le \frac{1}{|x^4|-1} \le \frac{2}{R^4} \to 0$$

so by the estimation lemma, the integral around the arc disappears.

$$\oint_C \frac{dz}{z^4+1} = \int_{-\infty}^\infty \frac{dx}{x^4+1} =2 \pi i \sum \operatorname*{Res}\frac{1}{z^4+1}$$

where the residues are of poles in the upper half plane. These poles are $z_1=e^{i\pi/4}$ and $z_2=e^{3i\pi/4}$. It follows that

$$b_1=\operatorname*{Res}_{z=z_1}\frac{1}{z^4+1} = -\frac{1}{4} e^{i\pi/4}$$ $$b_2=\operatorname*{Res}_{z=z_1}\frac{1}{z^4+1} = -\frac{1}{4} e^{3 i\pi/4}$$

Then

$$\int_{-\infty}^\infty \frac{dx}{x^4+1} =2 \pi i (b_1+b_2) = \frac{\pi}{\sqrt{2}}$$

and finally

$$\int_0^\infty \frac{dx}{x^4+1} = \frac{\pi}{2\sqrt{2}}$$

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  • $\begingroup$ You should at least say why integral along the arc disappears. $\endgroup$ – xyzzyz May 14 '13 at 20:41
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    $\begingroup$ @xyzzyz It's not difficult to show; I leave it as an exercise. $\endgroup$ – Argon May 14 '13 at 20:42
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    $\begingroup$ Yeah, but if you know how to do this, then you also know how to use residue theorem to calculate integrals like this. :) $\endgroup$ – xyzzyz May 14 '13 at 20:43
  • $\begingroup$ @xyzzyz Updated. $\endgroup$ – Argon May 14 '13 at 21:06
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This may be done using residue theory. Consider

$$\oint_C \frac{dz}{1+z^4}$$

where $C$ is a closed contour that spans the perimeter of the quarter-circle in the 1st quadrant (i.e., the positive real and positive imaginary quarter plane), of radius $R$. As $R \to \infty$, the integral over the circular arc vanishes, and we are left with this contour integral being equal to

$$(1-i) \int_0^{\infty} \frac{dx}{1+x^4}$$

This integral is equal to $ i 2 \pi$ times the residue at the pole $z=e^{i \pi/4}$. Thus

$$(1-i) \int_0^{\infty} \frac{dx}{1+x^4} = \frac{i 2 \pi}{4 e^{i 3 \pi/4}}$$

which means that

$$\int_0^{\infty} \frac{dx}{1+x^4} =\frac{\pi}{2 \sqrt{2}}$$

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make a change of variable $ x=y^{1/4} $ and use the identity

$$ \int_{0}^{\infty}\frac{t^{s-1}}{x+1}dt= \frac{\pi}{sin(\pi s)}$$

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