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