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What I tried was I took $$\tan^{-1}(x) =t, $$ but I got terms like $$\cos(\tan(t))$$ which I don't know what to do with. So please guide me into solving this..

Jyrki Lahtonen
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satyatech
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    Problems like this make students hate calculus. Good god. I'm sure there's a relatively simple solution, but.. a problem like this is needlessly complicated for a calculus student. – Cameron Williams Feb 26 '17 at 05:10
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    Sir I can't write inverse functions in mathjax please help me @CameronWilliams – satyatech Feb 26 '17 at 05:12
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    You were pretty close! $\rm\LaTeX$ requires curly brackets for exponents with more than one character, e.g. \tan^{-1}(x) to get $\tan^{-1}(x)$, but you could just do \sin^2(x) for $\sin^2(x)$. – Cameron Williams Feb 26 '17 at 05:14
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    In the picture, is $tan^{-1}x = arctan(x)$ or $cotx$? – mrnovice Feb 26 '17 at 05:15
  • @mrnovice. I think that you are right. It should be $\cot(x)$. Even in this case, it is a monster. – Claude Leibovici Feb 26 '17 at 05:18
  • If $\tan^{-1}(x)$ means the arctangent function, then Mathematica choked on this integral, so I very much doubt it has a closed form solution. But if $\tan^{-1}(x)$ means $\frac{1}{\tan(x)}=\cot(x)$, then Mathematica did find an antiderivative for it, and even only in elementary functions, although it's not a pretty one. – zipirovich Feb 26 '17 at 05:19
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    But I am really sorry mate it is in fact arctanx @zipirovich@mrnovice – satyatech Feb 26 '17 at 05:20
  • Holy bejesus. This is the answer Wolfram Alpha gives in the case of $\tan(x)^{-1}$. It hung on the case of it being $\arctan(x)$. – Cameron Williams Feb 26 '17 at 05:21
  • Anyone got wolfram alpha pro - my computation time was exceeded :(? http://www.wolframalpha.com/input/?i=integral+of+(sin%5E3x)%2F((cos%5E4x%2B3cos%5E2x%2B1)arctanx(secx%2Bcosx))+dx – mrnovice Feb 26 '17 at 05:23
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    Got the same on the app @mrnovice. Abandon hope ye who may enter here. – Cameron Williams Feb 26 '17 at 05:27
  • I have got Wolfram alpha pro from last 3 mins it is calculating .....@mrnovice – satyatech Feb 26 '17 at 05:27
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    Where did you get this problem from? – mrnovice Feb 26 '17 at 05:33
  • From my textbook for iitjee – satyatech Feb 26 '17 at 05:37
  • I don't know what iitjee is, but why would it give you such a problem? Is there a typo somewhere? – mrnovice Feb 26 '17 at 05:42
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    Not at all ,these type of questions are reputed ,,don't worry about the accuracy of this question ,dude,just think of answering it.....@mrnovice – satyatech Feb 26 '17 at 05:47
  • Don't be pushy. The reason for asking is that, as stated, this question seems impossible. – Cameron Williams Feb 26 '17 at 05:48
  • @satyatech Please upload a picture of the problem from your textbook directly. Probably you made a mistake while copying it here. –  Feb 26 '17 at 06:10
  • the site http://www.integral-calculator.com/ gives this result, I don't what the comma means ,$\ln\left(\operatorname{arctan2}\left(\cos^2\left(x\right)+1, \cos\left(x\right)\right)\right)$ – Vikram Feb 26 '17 at 06:24
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    @Vikram $\operatorname{arctan2}(x,y)$ is the polar angle coordinate of the point $(x,y)$. So when $x>0$ and $y>0$ we have $\operatorname{arctan2}(x,y)=\arctan(y/x).$ With other sign combos you may have to adjust by an integer multiple of $\pi$ (depending on the details of the definition). – Jyrki Lahtonen Feb 26 '17 at 06:35

1 Answers1

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Hint :
$$t = \arctan \left ( \sec x + \cos x \right ) ,\\dt = \frac{\sin ^{3}xdx}{\cos^{4}x+ 3\cos ^{2}x + 1 }\\ \int \frac{dt}{t} = \log_{e} \arctan \left ( \sec x + \cos x \right ) + C$$

JayN
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