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

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. – J. M. Oct 17 '11 at 14:04
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 $\frac{\pi}{4}$). – gaurav Oct 17 '11 at 14:58
@gaurav, Indeed, the limit should be $\frac{\pi}{4}$ in the LHS as well. – Tapu Oct 17 '11 at 15:06
@Swapan How? Can you please elaborate? If the limit is correct then ${}^1/_2$ shouldn't be there! Sorry. I'm confused. – gaurav Oct 17 '11 at 15:16
Oh.. I think I got it. – gaurav Oct 17 '11 at 15:19

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 $I=v(b)=\frac12(1-\log2)$.
You should know that $\cos\frac\pi4=\frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$. So $$\ln \cos\frac\pi4=\ln\frac{1}{\sqrt{2}} = -\ln\sqrt{2} = -\ln\left(2^{1/2}\right)= -\frac12\ln 2.$$