So, from here $$\int \frac{\sin(x)}{3\cos^3(x)+\sin^2(x)\cdot\cos(x)} dx$$ I divided by cos(x) and I got $$\int \frac{\tan(x)}{2\cos^2(x)+1} dx$$ But I'm stuck here. I tried to substitute $t=\cos(x)$

$$\int \frac{-1}{t\cdot(2t^2+1)} dt$$

Any help would be greatly appreciated.

  • 3
    $\begingroup$ Try $s=t^2$. (And please add $dx$ at the three places where it belongs.) $\endgroup$ – Did Jun 13 '12 at 17:54
  • $\begingroup$ That is, after the substitution $t=\cos(x)$ that was mentioned in a previous version of the post. $\endgroup$ – Did Jun 13 '12 at 17:55
  • $\begingroup$ I am inclined to think that the solution I posted below is probably the most straightforward one unless you have one that uses some surprising "trick". $\endgroup$ – Michael Hardy Jun 13 '12 at 21:09

Alternate solution

$$\int \frac{\sin(x)}{3\cos^3(x)+\sin^2(x)\cos(x)} dx=\int \frac{\tan(x)}{2\cos^2(x)+1} dx= \int \frac{1}{\cos^2(x)} \frac{\tan(x)}{2+\sec^2(x)} dx$$

Thus, after $t= \tan(x)$ you get

$$\int \frac{t dt}{t^2+3} $$

  • $\begingroup$ I didn't think to divide for $$cos^2(x)$$ and put $$sec^2(x) = 1+tan^2(x)$$ Thank you for the reply! $\endgroup$ – HelloEveryone Jun 13 '12 at 18:11

$$ \int \frac{\sin(x)}{3\cos^3(x)+\sin^2(x)\cdot\cos(x)}\,dx = \int \frac{1}{3\cos^3(x)+\sin^2(x)\cdot\cos(x)}\,\Big(\sin x \,dx\Big) $$ $$ = \int \frac{1}{3\cos^3(x)+(1-\cos^2 x)\cdot\cos(x)}\,\Big(\sin x \,dx\Big) = \int \frac{1}{3u^3 + (1-u^2)u}\,(-du) $$ Then use partial fractions.

Later edit in response to comments: $$ \int \frac{1}{3u^3 + (1-u^2)u}\,(-du) = \int\frac{-du}{u(2u^2 + 1)} = \int \frac{A}{u} + \frac{Bu+C}{2u^2+1} \, du $$ Two logarithms plus an arctangent.

  • $\begingroup$ I don't think it's possible to use partial fractions at that point. $\endgroup$ – Gigili Jun 13 '12 at 19:50
  • 1
    $\begingroup$ @Gigili : Fortunately you're mistaken; it's quite simple. I've added it to my answer. $\endgroup$ – Michael Hardy Jun 13 '12 at 21:04
  • $\begingroup$ Done. It's two logarithms plus an arctangent. $\endgroup$ – Michael Hardy Jun 13 '12 at 21:07
  • $\begingroup$ Well, now that you simplified it ... $\endgroup$ – Gigili Jun 13 '12 at 21:09

From the last integral, use $\frac{1}{t(2t^2+1)}=\frac{1}{t}-\frac{2t}{2t^2+1}$. Now, you have: $$\int \frac{1}{t\cdot(2t^2+1)} \, \mathrm{d}t=\int \frac{1}{t} \, \mathrm{d}t-\int \frac{2t}{2t^2+1} \, \mathrm{d}t=\ln|t|-\frac{1}{2}\ln|2t^2+1|+C$$

  • $\begingroup$ Thank you! You reach that point by usinf partial fraction, am I right? $\endgroup$ – HelloEveryone Jun 13 '12 at 18:09
  • $\begingroup$ @HelloEveryone: Yes, I did. $\endgroup$ – Dennis Gulko Jun 13 '12 at 18:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.