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Can someone explain why when going from this step:

$$\frac23\int \frac{\sin u}{\cos u}du$$

where substituting $s=\cos u$ and $ds = -\sin u\,du$ produces

$$\frac23\int -\frac1s ds$$

My work shows it to be this from the substitution of $s$ for $\cos u$ and $ds = -\sin u\,du$

$$\frac23 \int \frac{(\sin u)(-\sin u)}{s}ds= \frac23 \int -\frac{\sin^2 u}{s}ds$$

Im not sure how it is correctly reduced to $-1$.

Riley
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user04445
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  • I don't mean to be rude but I would think that, if you are taking a course in which you are expected to do a problem like this, you must have taken a trigonometry course previously. And one of the most basic "trig identities" is "$sin^2(\theta)+ cos^2(\theta)= 1$" from which $sin^2(\theta)= 1- cos^2(\theta)$. – user247327 Oct 09 '17 at 22:00
  • Notice that $ds=-\sin u du$, so $du=\frac{-ds}{\sin u}$ – Isai Dávila Cuba Oct 09 '17 at 22:02
  • I have considered that. But even still, if we substitute -(1-cos^ u) for -sin u, it would appear as 2/3 ∫ (1-cos ^2 u(-sinu))/ (s) ds, this is still not -1/s. If we distribute ds --> (-sinu - cos^2u) / s =/= -1/s – user04445 Oct 09 '17 at 22:05
  • That, perhaps, is not exactly what you are asking. You say you have $\frac{2}{3}\int \frac{sin(u)}{cos(u)}du$. Yes, if you let $s= cos(u)$ then $ds= -sin(u)du$. So $\frac{sin(u) du}{cos(u}= \frac{ds}{-s}$. You seem to have left "cos(u)" in the numerator instead of using it with "cos(u)du . You could also think "since $-sin(u)du= ds$ then $du= -\frac{1}{sin(u)}ds$. So \frac{sin(u)}{cos(u)}du= \frac{sin(u)}{s}\left(-\frac{1}{sin(u)} ds= -\frac{1}{s}ds$ That is you do not have those two "sin(u)"s multiplying. One is in the denominator so they cancel. – user247327 Oct 09 '17 at 22:07
  • When you've $\int \frac{f'(u)}{f(u)}du$, using the substitution $s=f(u)$ one has using the rule $d(f(u))=f'(u)du$ (that holds for good functions), that $1\cdot ds=d(s)=d(f(u))=f'(u)du$ and you can deduce $$\int \frac{1}{f(u)}f'(u)du=\int \frac{1}{s}ds$$ Good luck. –  Oct 09 '17 at 22:11

2 Answers2

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So you have:

$$I=\frac23\int{\frac{\sin{u}}{\cos{u}}d{u}}$$

Let:

$$ s=\cos{u}$$

Therefore: $$ d{s}=-\sin{u}\ d{u}$$

Now $\cos{u}$ is in the denominator so: $$I=\frac23\int-\frac{d{s}}{s}$$

Paul Evans
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You have$$I=\int\frac{\sin u}{\cos u}\,\mathrm du$$$$s=\cos u$$$$\mathrm ds=-\sin u \,\mathrm du$$So$$\implies I=\int\frac{\sin u}{\cos u}\,\mathrm du=\int-\frac{-\sin u\,\mathrm du}{\cos u}=\int-\frac{\mathrm ds}{s}=\int-\frac1s\,\mathrm ds$$

John Doe
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