Evaluate the integral

Question

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.

Answer 1

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?

Answer 2

$$\int\limits_0^\pi {\frac{{\cos x}}{{\sin {x^3} + \sin x + 4}}} dx % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8qCaeaada % WcaaqaaiGacogacaGGVbGaai4CaiaadIhaaeaaciGGZbGaaiyAaiaa % c6gacaWG4bWaaWbaaSqabeaacaaIZaaaaOGaey4kaSIaci4CaiaacM % gacaGGUbGaamiEaiabgUcaRiaaisdaaaaaleaacaaIWaaabaGaeqiW % dahaniabgUIiYdGccaWGKbGaamiEaaaa!4BC6! $$ 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 $$\int\limits_0^0 {\frac{{\cos x}}{{\sin {x^3} + \sin x + 4}}} dx = 0 % MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8qCaeaada % WcaaqaaiGacogacaGGVbGaai4CaiaadIhaaeaaciGGZbGaaiyAaiaa % c6gacaWG4bWaaWbaaSqabeaacaaIZaaaaOGaey4kaSIaci4CaiaacM % gacaGGUbGaamiEaiabgUcaRiaaisdaaaaaleaacaaIWaaabaGaaGim % aaqdcqGHRiI8aOGaamizaiaadIhacqGH9aqpcaaIWaaaaa!4C83! $$

Answer 3

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).