## A Tricky Exponent
*March 25, 2011*

*Posted by Saketh in : calculus , trackback*

Determine the exact value ofÂ $$\int^{\frac{\pi}{2}}_0 \frac{1}{1+(\tan{x})^{\sqrt{2}}} \,dx$$

**Solution**

Let $$I\left(\alpha\right)=\int^{\frac{\pi}{2}}_0 \frac{1}{1+\left(\tan{x}\right)^{\alpha}} \,dx$$

We shall exploit the symmetry of the integrand about $\frac{\pi}{4}$. Consider $$\int^{\frac{\pi}{2}}_0 \frac{1}{1+\left(\tan\left({\frac{\pi}{2}-x}\right)\right)^{\alpha}} \,dx$$

Notice that this is equal to our original integral, a fact which can be seen either by performing a $u$-substitution with $u=\frac{\pi}{2}-x$ or by observing that this new integral effectively “passes over” the same x-values as the original.

Simplifying the integrand, we get $$\frac{1}{1+\left(\tan\left({\frac{\pi}{2}-x}\right)\right)^{\alpha}}=\frac{1}{1+\frac{\left(\sin\left({\frac{\pi}{2}-x}\right)\right)^{\alpha}}{\left(\cos\left({\frac{\pi}{2}-x}\right)\right)^{\alpha}}}=\frac{1}{1+\frac{\left(\cos\left(x\right)\right)^{\alpha}}{\left(\sin\left(x\right)\right)^{\alpha}}}=\frac{\left(\tan\left(x\right)\right)^{\alpha}}{1+\left(\tan\left(x\right)\right)^{\alpha}}$$

Thus, $$I\left(\alpha\right)=\displaystyle\int^{\frac{\pi}{2}}_0 \frac{1}{1+\left(\tan{x}\right)^{\alpha}} \,dx = \displaystyle\int^{\frac{\pi}{2}}_0 \frac{\left(\tan\left(x\right)\right)^{\alpha}}{1+\left(\tan{x}\right)^{\alpha}} \,dx$$

and $$2I\left(\alpha\right)=\displaystyle\int^{\frac{\pi}{2}}_0 \frac{1}{1+\left(\tan{x}\right)^{\alpha}} \,dx + \displaystyle\int^{\frac{\pi}{2}}_0 \frac{\left(\tan\left(x\right)\right)^{\alpha}}{1+\left(\tan{x}\right)^{\alpha}} \,dx = \displaystyle\int^{\frac{\pi}{2}}_0 \,dx$$

So we can conclude that $2I\left(\alpha\right)=\frac{\pi}{2}$, and thus $I\left(\alpha\right)=\frac{\pi}{4}$. Interestingly, this result is independent of the value of the exponent $\alpha$.

Our original integral, $I\left(\sqrt{2}\right)$, therefore has a value of $\boxed{\frac{\pi}{4}}$.

## Comments»

hey look it’s albert’s favorite integral