Is there another way to solve an integral $$\int \frac{\sin^4(x)}{1+\cos^2(x)}\ dx$$ without the substitution $y=\tan\left(\frac{x}{2}\right)$?

$\large \int \frac{\sin^3(x)}{1+\cos^2(x)}\ dx$ is easily solved using the substitution $y=\cos(x)$. What if the power of sine is even?

  • $\begingroup$ $y = \tan x$ and $y = \cot x$ should also work. $\endgroup$
    – an4s
    Commented Jul 9, 2020 at 19:53
  • $\begingroup$ You may solve it numerically. But i am sure you want an analytic solution. $\endgroup$
    – ConvexHull
    Commented Jul 9, 2020 at 19:55
  • 1
    $\begingroup$ @ConvexHull: Huh? How can you solve that indefinite integral "numerically"?!? $\endgroup$ Commented Jul 9, 2020 at 20:02
  • $\begingroup$ You may always find an anti-derivative (or indefinite integral) approximately if you replace the integration kernel with a simpler expression, e.g. with a polynomial expansion of arbitrary order, before you apply the integration. This is often done with analytical !not! integrate able functions, e.g. the error function. $\endgroup$
    – ConvexHull
    Commented Jul 9, 2020 at 20:08
  • $\begingroup$ @DavidG.Stork Simply spoken, approximate $\frac{\sin^4(x)}{1+\cos^2(x)} \approx \sum_i^N a_i \Phi_i(x)$. With high $N$ you will get a quasi exact solution up to round of errors. $\endgroup$
    – ConvexHull
    Commented Jul 9, 2020 at 21:27

6 Answers 6



Bioche's rules suggest to use the substitution $$t=\tan x,\quad \mathrm d x=\frac{\mathrm dt}{1+t^2}.$$ Indeed, as $\cos^2x=\frac 1{1+t^2}$, $\:\sin ^2x=\frac{t^2}{1+t^2}$, one obtains $$\int\frac{\sin^4(x)}{1+\cos^2(x)}\,\mathrm dx=\int\frac{\frac{t^4}{(1+t^2)^2}}{1+\frac1{1+t^2}}\,\frac{\mathrm dt}{1+t^2}= \int\frac{t^4\,\mathrm dt}{(1+t^2)^2(2+t^2)}$$ which is comptuted with a decomposition into partial fractions: $$\frac{t^4}{(1+t^2)^2(2+t^2)}=\frac{At+B}{1+t^2}+\frac{Ct+D}{(1+t^2)^2}+\frac{Et+F}{2+t^2}.$$



$${I_n = \int\frac{\sin^{2n}(x)}{1 + \cos^2(x)}dx}$$



If you expand this, you get

$${=I_{n-1} - \int \cos^2(x)\frac{\sin^{2n-2}(x)}{1+\cos^2(x)}dx=I_{n-1}-\int \sin^{2n-2}(x) - \frac{\sin^{2n-2}(x)}{1+\cos^2(x)}dx}$$


$${I_{n}=2I_{n-1} - \int \sin^{2n-2}(x)dx}$$

Now define ${S_n = \int\sin^{2n}(x)dx}$. Then

$${S_{n}=\int \sin^2(x)\sin^{2n-2}(x)dx=S_{n-1}-\int \cos^2(x)\sin^{2n-2}(x)dx}$$

On the rightmost integral, using integration by parts yields

$${\int\cos^2(x) \sin^{2n-2}(x)dx=\frac{\cos(x)\sin^{2n-1}(x)}{2n-1}+\frac{1}{2n-1}\int \sin^{2n}(x)dx}$$

So overall

$${\Rightarrow S_n = S_{n-1}-\frac{\cos(x)\sin^{2n-1}(x)}{2n-1} - \frac{1}{2n-1}S_n}$$

And so

$${\left(\frac{2n}{2n-1}\right)S_n = S_{n-1} - \frac{\cos(x)\sin^{2n-1}(x)}{2n-1}}$$

$${\Rightarrow S_n = \frac{(2n-1)S_{n-1}}{2n} - \frac{\cos(x)\sin^{2n-1}(x)}{2n}}$$

Now you have two recursive relations that will help you compute the integral for higher even powers of ${\sin(x)}$:

$${I_n = 2I_{n-1} - S_{n-1}}$$

$${S_{n} = \frac{(2n-1)S_{n-1}}{2n} - \frac{\cos(x)\sin^{2n-1}(x)}{2n}}$$


HINT: $$\int \frac{\sin^4(x)}{1+\cos^2(x)}\ dx=\int \frac{(1-\cos^2(x))^2}{1+\cos^2(x)}\ dx$$ $$=\int \frac{(1+\cos^2(x))^2-4(1+\cos^2(x))+4}{1+\cos^2(x)}\ dx$$ $$=\int \left(1+\cos^2(x)-4+\frac{4}{1+\cos^2(x)}\right)\ dx$$ $$=\int \left(\frac{1+\cos(2x)}{2}-3+\frac{4\sec^2x}{\tan^2(x)+2}\right)\ dx$$

  • 1
    $\begingroup$ Will this work for every even power of sine? $\endgroup$
    – eMathHelp
    Commented Jul 9, 2020 at 20:06
  • 6
    $\begingroup$ for higher even powers, the conversion $\sin^2x\to \cos^2x$ is possible but becomes much lengthier and complicated $\endgroup$ Commented Jul 9, 2020 at 20:10
  • 3
    $\begingroup$ For higher even powers of sine, you can use a recurrence relation combined with Harish' answer, as you can see below @AndrewFount $\endgroup$ Commented Jul 9, 2020 at 20:36


You can use $\sin(x) = \tan(x) / \sec(x)$ and $\cos(x) = 1/\sec(x)$.

and then it will bring you to Partial Fractions Decomposition.


$$\sin^4x=(\sin^2x+1)(\sin x+\sqrt{2})(\sin x-\sqrt{2})+(2+\sin^2x)$$ $$1+\cos^2x=(\sqrt{2}+\sin x)(\sqrt{2}-\sin x)$$ and so your integral becomes: $$\int\frac{\sin^4x}{1+\cos^2x}dx=\int\frac{(\sin^2x+1)(\sin x+\sqrt{2})(\sin x-\sqrt{2})+(2+\sin^2x)}{(\sqrt{2}+\sin x)(\sqrt{2}-\sin x)}dx=-\int1+\sin^2xdx+\int\frac{2+\sin^2x}{2-\sin^2x}dx$$


If you enjoy special functions, using $t=\tan(x)$ $$I_n=\int \frac{\sin^n(x)}{1+\cos^2(x)}\ dx=\int \left(\frac{t}{\sqrt{t^2+1}}\right)^n\frac{dt}{t^2+2}$$

$$I_n=\frac {t^{n+1}}{2(n+1)}\,F_1\left(\frac{n+1}{2};\frac{n}{2},1;\frac{n+3}{2};-t^2,-\frac{t^2}{2}\right)$$ where appears the Appell hypergeometric function of two variables.


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