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I am trying to do this problem completely on my own but I can not get a proper answer for some reason

$$\begin{align} \int \cos^2 x \tan^3x dx &=\int \cos^2 x \frac{ \sin^3 x}{ \cos^3 x}dx\\ &=\int \frac{ \cos^2 x\sin^3 x}{ \cos^3 x}dx\\ &=\int \frac{ \sin^3 x}{ \cos x}dx\\ &=\int \frac{ \sin^2 x \sin x}{ \cos x}dx\\ &=\int \frac{ (1 -\cos^2 x) \sin x}{ \cos x}dx\\ &=\int \frac{ (\sin x -\cos^2 x \sin x) }{ \cos x}dx\\ &=\int \frac{ \sin x -\cos^2 x \sin x }{ \cos x}dx\\ &=\int \frac{ \sin x }{ \cos x}dx - \int \cos x \sin x dx\\ &=\int \tan x dx - \frac{1}{2}\int 2 \cos x \sin x dx\\ &=\ln|\sec x| - \frac{1}{2}\int \sin 2x dx\\ &=\ln|\sec x| + \frac{\cos 2x}{4} + C \end{align}$$

This is the wrong answer, I have went through and back and it all seems correct to me.

  • 0
    It isn't a constant, I am suppose to get $\frac{1}{2}\cos^2 x - \ln|\cos x| + C$2012-06-08
  • 10
    Same function, different expression. Note that $$\ln|\sec x| = \ln\left|\frac{1}{\cos x}\right| = \ln\left(|\cos x|^{-1}\right) = -\ln|\cos x|.$$ And $$\cos 2x = \cos^2x - \sin^2x = \cos^2x-(1-\cos^2x) = 2\cos^2x - 1$$so $$\frac{\cos 2x}{4}+C = \frac{2\cos^2x}{4}-\frac{1}{4}+C = \frac{1}{2}\cos^2x + C'.$$2012-06-08
  • 4
    You're making good progress I see.2012-06-08
  • 1
    Alright, now I would like to know why the first integral gives $\ln \sec x$ instead of $-\ln \cos x$.2012-06-08
  • 0
    I think that was just an error in typing out the question2012-06-08
  • 1
    @Phira: Because you can bring the $-1$ up.2012-06-08

7 Answers 7