I came across this integral while studying indefinite integrals.

So far I have had many unsuccessful attempts which include - trying by parts - but I could not find a way to proceed with it.

I even tried dividing both numerator and denominator by $x^4$ to yield $$\int \frac {(1+x^{-2})(1+2x^{-2})}{( \cos x+ \frac \sin x)^4}dx.$$ But I am still not able to move forward.

I even tried to cheat a little bit by taking derivative of options, still nothing.

And many more..

So the problem still stands. Can someone tell me how to proceed?

(Note: this is a problem from very elementary calculus course so no contour integrals, no multivariable and such stuff, however I think differentiation under integration would be fine. Also it would help if answers are one of those present in options (see image).)

Edit: Thanks to comments now I know answer is C but I was wondering if anyone could show me a straightforward way to do it thanks! Edit 2 : turns out B is also correct.

  • $\begingroup$ Try derivatives of each option. If needed, plot to see which is off by vertical shift from starting integrand. $\endgroup$
    – coffeemath
    Apr 5 '20 at 2:51
  • $\begingroup$ As coffeemath has suggested, try taking the derivatives of each option. Instead of doing that manually, there are online resources to help, with WolframAlpha being a common one which is used. However, one interesting thing I've noticed, which might help but I'm not sure as I didn't follow through on it on checking it further, is that if $f(x) = x\sin x$, then $f'(x) = x\cos x + \sin x$. $\endgroup$ Apr 5 '20 at 2:55
  • $\begingroup$ @John Omielan Thanks! Now I know answer is C. $\endgroup$ Apr 5 '20 at 3:02
  • $\begingroup$ @coffeemath thanks! I know the answer now. $\endgroup$ Apr 5 '20 at 3:08

Let $f_n(x) = \left( \frac {\cos x - x \sin x}{x\cos x + \sin x} \right)^n$ and note that,

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

Then, for $n=1$ and $n=3$, we have respectively, $$\frac{df_1(x)}{dx} = -\frac{x^2+2}{(x\cos x + \sin x)^{2}}$$ $$\frac{df_3(x)}{dx}=- \frac{3(\cos x - x \sin x)^2(x^2+2)}{(x\cos x + \sin x)^{4}}$$

which leads to

$$\frac {(1+x^2)(2+x^2)}{(x \cos x+\sin x)^4} = -\frac{d}{dx}\left(\frac13f_3(x)+f_1(x)\right)$$


$$\begin{align} & \int \frac {(1+x^2)(2+x^2)}{(x \cos x+\sin x)^4}dx \\ & =-\frac13f_3(x)-f_1(x) \\ & =-\frac13 \left( \frac {\cos x - x \sin x}{x\cos x + \sin x} \right)^3 - \frac {\cos x - x \sin x}{x\cos x + \sin x} \\ & =-\frac13 \left( \frac {1- x \tan x}{x + \tan x} \right)^3 - \frac {1- x \tan x}{x + \tan x} + C\\ \end{align}$$

  • $\begingroup$ Wow! That was amazing. Can I ask how did you came up with the fact that this $$f_n(x) = \left( \frac {\cos x - x \sin x}{x\cos x + \sin x} \right)^n$$ might help? $\endgroup$ Apr 5 '20 at 4:12
  • $\begingroup$ @HrishabhNayal - I knew about this general expression; I have seen it. It helps produce high inverse power $\endgroup$
    – Quanto
    Apr 5 '20 at 4:18
  • $\begingroup$ @Quanto !Very nice indeed! $\endgroup$
    – Z Ahmed
    Apr 5 '20 at 4:50

Let $$I=\int\frac{(1+x^2)(2+x^2)}{(x\sin x+\cos x)^4}dx$$

We can write $x\sin x+\cos x=\sqrt{1+x^2}\cos(x-\alpha)$

Where $\displaystyle \sin \alpha=\frac{x}{\sqrt{x^2+1}}$ and $\displaystyle \cos \alpha=\frac{1}{\sqrt{x^2+1}}$ and $\tan \alpha=x\Rightarrow \alpha=\tan^{-1}(x)$

So $$I =\int\frac{2+x^2}{1+x^2}\sec^4(x-\tan^{-1}(x))dx$$

  • $\begingroup$ Is that a hint? If so I am unable to follow it to answer can you elaborate further? $\endgroup$ Apr 5 '20 at 3:36
  • $\begingroup$ Hey! I got it now. Thanks! Now I kniw B is also correct. :) $\endgroup$ Apr 5 '20 at 7:15

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.