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I have the following integral

$$\int_{0}^{\pi / 2} \mathrm{d} x \,\, \frac{\cos^{3}(x/2) - \cos^{4} (x)}{\sin^{2} (x)}$$

My current solution is to use $$v = \tan \left(\frac{x}{4}\right)$$ to obtain a rational function of $v$, and integrate this.

Is there a more practical / clever way of doing it?

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  • $\begingroup$ Where did you see this integral? $\endgroup$ – Frpzzd Nov 19 '18 at 22:44
  • $\begingroup$ Comes up in my work. I like the Weierstrass trick as it is a blanket method, but am looking for a better way to approach it $\endgroup$ – Gordon Nov 19 '18 at 22:46
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While the substitution $v=\tan(x/4)$ would work, I don't think it's a good method in this case, which just requires elementary antiderivatives.

First, split the fraction and notice that $$ \int\frac{\cos^4x}{\sin^2x}\,dx=\int\left(\frac{1}{\sin^2x}-2+\sin^2x\right)\,dx =-\cot x-2x+\frac{1}{2}(x-\sin x\cos x) $$ This leaves the other piece: $$ \int\frac{\cos^3(x/2)}{\sin^2x}\,dx=[t=x/2]=2\int\frac{\cos^3t}{\sin^22t}\,dt =\frac{1}{2}\int\frac{\cos t}{\sin^2t}\,dt=-\frac{1}{2\sin t} $$ Thus the integral is $$ -\frac{1}{2\sin(x/2)}+\cot x+2x-\frac{1}{2}(x-\sin x\cos x) $$ You can also note that $$ \cot x-\frac{1}{2\sin(x/2)}=\frac{\cos x}{2\sin(x/2)\cos(x/2)}-\frac{1}{2\sin(x/2)}= \frac{(\cos(x/2)-1)(2\cos(x/2)+1)}{2\sin(x/2)} $$ which has limit $0$ for $x\to0$.

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$$\frac{\cos^3\dfrac x2}{\sin^2x}=\frac{\cos\dfrac x2}{2\sin^2\dfrac x2}$$ is immediately integrable.

And in

$$\frac{\cos^4x}{\sin^2x}=\frac1{\sin^2x}-2+\sin^2x,$$ the first two terms are also immediate, and

$$\sin^2x=\frac{1-\cos2x}2.$$

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$$J=\int\frac{\cos^3(\frac x2)-\cos^4x}{\sin^2x}\mathrm{d}x$$ $$J=\int\frac{\cos^3(\frac x2)}{\sin^2x}\mathrm{d}x-\int\frac{\cos^4x}{\sin^2x}\mathrm{d}x$$ $$I=J\bigg|_0^{\pi/2}$$


$$I_1=\int\frac{\cos^3(\frac x2)}{\sin^2x}\mathrm{d}x$$ $u=x/2$: $$I_1=2\int\frac{\cos^3u}{\sin^22u}\mathrm{d}u$$ $$I_1=2\int\frac{\cos^3u}{4\sin^2u\cos^2u}\mathrm{d}u$$ $$I_1=\frac12\int\frac{\cos u}{\sin^2u}\mathrm{d}u$$ $t=\sin u$: $$I_1=\frac12\int\frac{\mathrm{d}t}{t^2}$$ $$I_1=-\frac1{2t}$$ $$I_1=-\frac1{2\sin(x/2)}$$


$$I_2=\int\frac{\cos^4x}{\sin^2x}\mathrm{d}x$$ $$I_2=\int\frac{\cos^2x(1-\sin^2x)}{\sin^2x}\mathrm{d}x$$ $$I_2=\int\cot^2x\ \mathrm{d}x-\int\cos^2x\ \mathrm{d}x$$ $$I_2=-\bigg(\cot x+\frac12\cos x\,\sin x+\frac32x\bigg)$$


$$I=\bigg(-\frac12\csc(x/2)+\cot x+\frac12\cos x\sin x+\frac32x\bigg)\bigg|_0^{\pi/2}$$ $$I=\frac{3\pi-2\sqrt{2}}4$$

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