I'm interested in the following problem:

Evaluate the indefinite integral$$I=\int\frac{\sin(x)+\cos(x)}{\sin^4(x)+\cos^2(x)}~dx$$

Here's what I did:

We note that $$I = \int\frac{\sin(x)}{\sin^4(x)+\cos^2(x)}~dx+\int\frac{\cos(x)}{\sin^4(x)+\cos^2(x)}~dx$$Now we evaluate each part separately. Let $\cos(x)=u$ and thus, $$I_1:=\int\frac{\sin(x)}{\sin^4(x)+\cos^2(x)}~dx=\int\frac{-du}{(1-u^2)^2+u^2}=-\int\frac{du}{1+u^4-u^2}$$For the second part, we substitute $\sin(x)=v$ and so, $$I_2:=\int\frac{\cos(x)}{\sin^4(x)+\cos^2(x)}~dx = \int\frac{dv}{v^4+1-v^2}=\int\frac{du}{u^4+1-u^2}$$ Now as, $I=I_1+I_2$, we conclude $$I=\int 0~ du = \text{Constant}$$However, we get a different answer by WolframAlpha.

So my question where am I wrong in above process?

Any help will be highly appreciated.

  • 4
    $\begingroup$ Well, write $\int 2 dx=\int (1+1)dx=\int dx+\int dx=\int dx-\int dy$ (taking $x=-y$). Now change $dy$ to $dx$ as you did and you have $\int 2 dx=0$. $\endgroup$
    – cqfd
    Sep 2, 2020 at 16:04
  • 2
    $\begingroup$ You simply forgot to revert to the original variables. $\endgroup$
    – Bernard
    Sep 2, 2020 at 16:04
  • 1
    $\begingroup$ After the integration in $u$, or $v,$ you have to substitute back $u=\cos x$ or $v=\sin x$, so you cannot simply add the two results $\endgroup$ Sep 2, 2020 at 16:05
  • $\begingroup$ @enzotib Oh I got it! Thank you so much :). $\endgroup$
    – Anand
    Sep 2, 2020 at 16:08

1 Answer 1


You can factorize $$ u^4-u^2+1=(u^2+\sqrt{3} u+1)(u^2-\sqrt{3} u+1), $$ so that the partial fraction decomposition is $$ \frac{1}{u^4-u^2+1}=\frac{1}{4 \sqrt{3}}\left[\frac{(2 u+\sqrt{3})+\sqrt{3}}{u^2+\sqrt{3} u+1}-\frac{(2 u-\sqrt{3})-\sqrt{3}}{u^2-\sqrt{3} u+1}\right], $$ and we have $$ \int\frac{1}{u^4-u^2+1}du=\frac{1}{4 \sqrt{3}}\log \left(\frac{u^2+\sqrt{3} u+1}{u^2-\sqrt{3} u+1}\right)+\frac{1}{2} \left[\arctan\left(2 u+\sqrt{3}\right)+\arctan\left(2u-\sqrt{3}\right)\right]+C. $$ The two integrals become $$ -\frac{1}{4 \sqrt{3}}\log \left(\frac{\cos ^2x+\sqrt{3} \cos x+1}{\cos ^2x-\sqrt{3} \cos x+1}\right)-\frac{1}{2} \left[\arctan \left(2 \cos x+\sqrt{3}\right)+\arctan\left(2 \cos x-\sqrt{3}\right)\right]+C, $$ and $$ +\frac{1}{4 \sqrt{3}}\log \left(\frac{\sin ^2x+\sqrt{3} \sin x+1}{\sin ^2x-\sqrt{3} \sin x+1}\right)+\frac{1}{2} \left[\arctan\left(2 \sin x+\sqrt{3}\right)+\arctan\left(2 \sin x-\sqrt{3}\right)\right]+C. $$

  • $\begingroup$ Thanks for you help! :) $\endgroup$
    – Anand
    Sep 3, 2020 at 7:56

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