definite integral

**jacks** · January 27th 2011, 06:16 AM

$\int_{0}^{\frac{\pi}{4}}xtanxdx$

**HallsofIvy** · January 27th 2011, 06:23 AM

Almost anytime you have x, or a power of x, multiplied by another function of x, you should think "integration by parts". If you let u= x, then du= dx so you have gotten rid of that x (if you have a power of x, setting u equal to that power of x reduces the power and repeated integration by parts will eliminate it altogether).
Of course, now dv= tan(x)dx. Do you know the anti-derivative of tan(x)?

If not, think of it as $\int \frac{sin(x)}{cos(x)}dx$ and use the substitution u= cos(x).

**Prove It** · January 27th 2011, 06:30 AM

Originally Posted by HallsofIvy

Almost anytime you have x, or a power of x, multiplied by another function of x, you should think "integration by parts". If you let u= x, then du= dx so you have gotten rid of that x (if you have a power of x, setting u equal to that power of x reduces the power and repeated integration by parts will eliminate it altogether).
Of course, now dv= tan(x)dx. Do you know the anti-derivative of tan(x)?

If not, think of it as $\int \frac{sin(x)}{cos(x)}dx$ and use the substitution u= cos(x).

Except in this case IBP reduces to needing to evaluate the disgusting integral $\displaystyle \int{-\ln{|\cos{x}|}\,dx}$ ...

**TheCoffeeMachine** · January 27th 2011, 08:38 AM

So by integration by parts, we have:

$\begin{aligned} \displaystyle \int_{0}^{\frac{\pi}{4}}x\tan{x}\;{dx} &= \biggl\|- x\ln(\cos{x})\biggl\|_{x = 0}^{x = \frac{\pi}{4}} +\int_{0}^{\frac{\pi}{4}}\ln\left(\cos{x}}\right)\ ;{dx} = \frac{1}{8}\pi\ln(2)+\int_{0}^{\frac{\pi}{4}}\ln\l eft(\cos{x}}\right)\;{dx}.\end{aligned}$

At this point, I recall that one of the definitions of Catalan's constant was $\displaystyle \int_{0}^{\frac{\pi}{4}}\ln\left(2\cos{t}\right)\; {dt} = \frac{1}{2}K.$

So I write $\begin{aligned}\displaystyle \int_{0}^{\frac{\pi}{4}}\ln\left(\cos{x}}\right) \;{dx} & = \int_{0}^{\frac{\pi}{4}}\ln\left(2\cos{x}}\right) \;{dx}-\int_{0}^{\frac{\pi}{4}}\ln(2)\;{dx} = \frac{1}{2}K -\frac{\pi}{4}\ln(2).\end{aligned}$

Thus $\displaystyle \int_{0}^{\frac{\pi}{4}}x\tan{x}\;{dx} = \frac{\pi}{8}\ln(2)+\frac{1}{2}K -\frac{\pi}{4}\ln(2) = \frac{1}{2}K-\frac{1}{8}\pi\ln(2).$

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