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I need some help in evaluating this integral:

$$\int {\frac{{\cos x}}{{{{\sin }^3}x + \sin x + 4}}dx} $$

I've tried using the substitution $u=\sin{x}$ but I ended up with a cubic polynomial in the denominator.

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  • $\begingroup$ Could you show us some of the work you have done so far? $\endgroup$ – user344249 Apr 1 '17 at 22:08
  • $\begingroup$ You had a strange code accompanying your integral. I have suppressed it because it can be viral. $\endgroup$ – Jean Marie Apr 1 '17 at 22:12
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    $\begingroup$ I tried the substitution u = sinx but I ended up with a cubic polynomial in the denominator. $\endgroup$ – user382662 Apr 1 '17 at 22:12
  • $\begingroup$ @user382662 I've edited your question to include your efforts in the question (I hope you don't mind). That way, people who read the question know exactly where you are stuck so then you are more likely to get better answers. I suggest that you also add the cubic polynomial you obtained as a result of the substitution. I will add an upvote if you do so. $\endgroup$ – projectilemotion Apr 1 '17 at 22:22
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    $\begingroup$ Have you tried the substitution $\cos x=(1-t^2)/(1+t^2)$, $\sin x=(2t)/(1+t^2)$ with $t=\tan(x/2)$ ? $\endgroup$ – Jean Marie Apr 1 '17 at 22:22
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You can evaluate this integral by an $u=\sin x$ substitution and partial fraction decomposition. The solution requires very ugly complex third degree polynomial roots and takes 3 lines to write down, though. Are you sure you wrote down the problem correctly?

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  • $\begingroup$ I have the same result when I used integral calculator, I'am sure the problem is correct because it was in my exam like that . $\endgroup$ – user382662 Apr 1 '17 at 22:18
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$$\int\limits_0^\pi {\frac{{\cos x}}{{\sin {x^3} + \sin x + 4}}} dx % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8qCaeaada % WcaaqaaiGacogacaGGVbGaai4CaiaadIhaaeaaciGGZbGaaiyAaiaa % c6gacaWG4bWaaWbaaSqabeaacaaIZaaaaOGaey4kaSIaci4CaiaacM % gacaGGUbGaamiEaiabgUcaRiaaisdaaaaaleaacaaIWaaabaGaeqiW % dahaniabgUIiYdGccaWGKbGaamiEaaaa!4BC6! $$ I'am sorry it was definite integral not indefinite,so there's no need to integrate ,because when we make the substitution $$u = \sin x % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyDaiabg2 % da9iGacohacaGGPbGaaiOBaiaadIhaaaa!3BCB! $$ $$du = \cos xdx % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaadw % hacqGH9aqpciGGJbGaai4BaiaacohacaWG4bGaamizaiaadIhaaaa!3E95! $$
$$\begin{array}{l}x = 0\\u = \sin (0) = 0\\x = \pi \\u = \sin (\pi ) = 0\end{array} % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWG4b % Gaeyypa0JaaGimaaqaaiaadwhacqGH9aqpciGGZbGaaiyAaiaac6ga % caGGOaGaaGimaiaacMcacqGH9aqpcaaIWaaabaGaamiEaiabg2da9i % abec8aWbqaaiaadwhacqGH9aqpciGGZbGaaiyAaiaac6gacaGGOaGa % eqiWdaNaaiykaiabg2da9iaaicdaaaaa!4ED5! $$ So, by definite integrals properties ,the integral becomes $$\int\limits_0^0 {\frac{{\cos x}}{{\sin {x^3} + \sin x + 4}}} dx = 0 % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8qCaeaada % WcaaqaaiGacogacaGGVbGaai4CaiaadIhaaeaaciGGZbGaaiyAaiaa % c6gacaWG4bWaaWbaaSqabeaacaaIZaaaaOGaey4kaSIaci4CaiaacM % gacaGGUbGaamiEaiabgUcaRiaaisdaaaaaleaacaaIWaaabaGaaGim % aaqdcqGHRiI8aOGaamizaiaadIhacqGH9aqpcaaIWaaaaa!4C83! $$

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  • $\begingroup$ Okay now let's look at this. $\int_0^\pi e^{\sin(x) } dx$ with your substitution you would say that the integral equals zero? But the function $e^{\sin x} $ is strict positive. So something goes wrong, don't you think the same? The problem in your argument is that the substitution $u=\sin(x) $ is NOT one-to-one for $x\in[0,\pi] $. You must try something else. $\endgroup$ – Shashi Apr 2 '17 at 9:28
  • $\begingroup$ However, you might still get zero. But the argument you wrote is not a valid argument in general. You have seen it in the example I provided. $\endgroup$ – Shashi Apr 2 '17 at 9:45
  • $\begingroup$ In your problem$$\int\limits_0^\pi {{e^{\sin x}}dx} $$ We can't simply make the substitution $$u = \sin x$$ because there's no $$\cos x$$ in the integral to make $$du = \cos xdx$$ $\endgroup$ – user382662 Apr 2 '17 at 11:30
  • $\begingroup$ One can surely substitute $\sin(x) $ without having $\cos(x) dx$. You can substitute whatever you want. The problem is that it does not always help. $\endgroup$ – Shashi Apr 2 '17 at 11:36
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For anyone who would like to see the method for evaluating the indefinite integral:

$\int{\frac{\mathrm{cos}x}{\mathrm{sin}^3x + \mathrm{sin}x + 4}}\mathrm{dx} = \int{\frac{\mathrm{du}}{u^3 + u + 4}}$

Which is by the substitution $u = \mathrm{sin}x$

Then let the roots of the polynomial be $r_1, r_2,$ and $r_3$. The fraction can then be split up:

$\frac{1}{u^3 + u + 4} = \frac{1}{(u-r_1)(u-r_2)(u-r_3)} = \frac{1}{(u-r_1)(r_1-r_2)(r_1-r_3)} + \frac{1}{(u-r_2)(r_2-r_1)(r_2-r_3)} + \frac{1}{(u-r_3)(r_3-r_1)(r_3-r_2)} $

Which can be integrated quite easily as:

$\int{(\frac{1}{(u-r_1)(r_1-r_2)(r_1-r_3)} + \frac{1}{(u-r_2)(r_2-r_1)(r_2-r_3)} + \frac{1}{(u-r_3)(r_3-r_1)(r_3-r_2)}})\mathrm{dx}\\= \frac{\mathrm{log}_e(u-r_1)}{(r_1-r_2)(r_1-r_3)} + \frac{\mathrm{log}_e(u-r_2)}{(r_2-r_1)(r_2-r_3)} + \frac{\mathrm{log}_e(u-r_3)}{(r_3-r_1)(r_3-r_2)} + \mathrm{C}\\= \frac{\mathrm{log}_e(\mathrm{sin}x-r_1)}{(r_1-r_2)(r_1-r_3)} + \frac{\mathrm{log}_e(\mathrm{sin}x-r_2)}{(r_2-r_1)(r_2-r_3)} + \frac{\mathrm{log}_e(\mathrm{sin}x-r_3)}{(r_3-r_1)(r_3-r_2)} + \mathrm{C}$

Now the last thing to do is to find $r_1, r_2,$ and $r_3$. Now, I didn't actually know how to do this, so I just looked at this which I used to find the roots as approximately:

$-1.38, 0.69 + 1.56i, 0.69 - 1.56i$.

If you want you can get the exact values by following the link (you can get the first root via this method, and then the next two can easily be deduced from a resulting quadratic).

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