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Find $\int\nolimits^{\frac{\pi}{4}}_{0} ( \tan^3{x} ) \space dx$ given $2\tan^3x = \frac{d}{dx}( \tan^2x+2\ln \cos x )$

$$\int\nolimits^{\frac{\pi}{4}}_{0} \tan^3{x} \space dx = \frac{1}{2}\left[\tan^2{x} + 2 \ln{\cos{x}}\right]^{\frac{\pi}{4}}_0$$

I could go on but $\cos{\frac{\pi}{4}}$ is a decimal. Answer is $\frac{1}{2}(1 - \ln{2})$. How do I simplify down to that

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    Uh, no. $\cos\frac{\pi}{4}$ is a very simple irrational number. You might want to recall the trigonometric functions of special angles...after which you can use the usual logarithm identities.2011-10-17
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    You wrote: $\displaystyle {\int_{0}^{^\pi/_2}} \tan^3 x \ dx$ $=\frac{1}{2} {\left({\tan^2 x +\ln \cos x}\right)}_{0}^{\pi/4}$ Which is incorrect, as I think. In RHS result upper limit should be $\pi/2$ (instead of $\frac{\pi}{4}$).2011-10-17
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    @gaurav, Indeed, the limit should be $\frac{\pi}{4}$ in the LHS as well.2011-10-17
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    @Swapan How? Can you please elaborate? If the limit is correct then ${}^1/_2$ shouldn't be there! Sorry. I'm confused.2011-10-17
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    Oh.. I think I got it.2011-10-17
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    You wrote $2\tan^3x = \frac{d}{dx}\tan^2x+2\ln \cos x$ when you seem to have meant $2\tan^3x = \frac{d}{dx}\left(\tan^2x+2\ln \cos x\right)$.2011-10-17
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    @jie In most beginning calculus courses, you might be expected to give "exact answers": no calculators! :-) (e.g. `(1-ln 2)/2`, instead of `0.1534`). Even when you need numerical answers, you might want to reserve the calculator for the last step, where you plug in values and simplify. So, you might want to remember some common values like trig ratios of angles like $\pi/4, \pi/3, \pi/6, 0, \pi$ and so on.2011-10-17
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    Ah, corrected that typo2011-10-18
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    It should be $\int^{\frac{\pi}{4}}_{0}$... not $\int^{\frac{\pi}{2}}_{0}$2011-10-18

2 Answers 2

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This seems to ask for the value of $I=\int\limits_a^bu(x)\,\mathrm dx$ knowing that $u(x)=v'(x)$ for every $x$, for certain values of $a$ and $b$ and certain functions $u$ and $v$. The answer is $I=v(b)-v(a)$.

Assume that $v(x)=\frac12\tan(x)^2+\log\cos(x)$, $a=0$ and $b=\frac\pi 4$. Then $v(a)=\frac12\times0+\log(1)=0$, $v(b)=\frac12\times1+\log\frac{\sqrt2}2$ and $\log\frac{\sqrt2}2=-\log\sqrt2=-\frac12\log2$, hence $I=v(b)=\frac12(1-\log2)$.

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You should know that $\cos\frac\pi4=\frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$. So $$\ln \cos\frac\pi4=\ln\frac{1}{\sqrt{2}} = -\ln\sqrt{2} = -\ln\left(2^{1/2}\right)= -\frac12\ln 2.$$