5

I attempted to integrate the following function from a practice problem in my Calculus textbook:

$$\displaystyle \int_{0}^{\frac{\pi}{2}}{\frac{1}{1+\tan^\sqrt{2}(x)}} \ {\rm d}x$$

I failed to find an indefinite integral, and I am assuming getting an indefinite integral is simply impossible. Using Wolfram|Alpha to estimate the definite integral, I got $0.785398$. I am assuming this is $\frac{\pi}{4}$, but I have no formal proof that this is the answer.

r9m
  • 17,938
Nicholas Pipitone
  • 1,192

2 Answers2

5

I finally figured it out due to a hint by Sameer Kailasa.

$$\int_{0}^{\frac{\pi}{2}}{\frac{1}{1+\tan^\sqrt{2}(x)}} \ dx$$ $$u=\frac{\pi}{2}-x \implies du=-dx$$ $$= -\int_{\frac{\pi}{2}}^{0}{\frac{1}{1+\cot^\sqrt{2}(x)}} \ dx = \int_{0}^{\frac{\pi}{2}}{\frac{\tan^\sqrt{2}(x)}{\tan^\sqrt{2}(x)+1}} \ dx $$ $$=\frac{1}{2}\int_{0}^{\frac{\pi}{2}}{\frac{1}{1+\tan^\sqrt{2}(x)}+\frac{\tan^\sqrt{2}(x)}{\tan^\sqrt{2}(x)+1}} \ dx$$ $$\frac{1}{2}\int_{0}^{\frac{\pi}{2}} \frac{1+\tan^\sqrt{2}(x)}{\tan^\sqrt{2}(x)+1} \ dx =\frac{1}{2} [{x}]_{0}^{\frac{\pi}{2}} = \frac{\pi}{4}$$

Nicholas Pipitone
  • 1,192
3

The $\sqrt{2}$ is a complete red herring. In fact consider

$$I=\int_0^{\frac{\pi}{2}}\frac{1}{1+\tan^{\alpha}(x)}\,dx$$

where $\alpha$ is any nonnegative real number. Then we have

\begin{align} I&=\int_0^{\frac{\pi}{2}}\frac{1}{1+\tan^{\alpha}(x)}\,dx \\&=\int_0^{\frac{\pi}{2}}\frac{1}{1+\frac{\sin^{\alpha}(x)}{\cos^\alpha(x)}}\,dx \\&=\int_0^{\frac{\pi}{2}}\frac{\cos^{\alpha}(x)}{\cos^{\alpha}(x)+\sin^{\alpha}(x)}\,dx \\&=\int_0^{\frac{\pi}{2}}\frac{\sin^{\alpha}(y)}{\cos^{\alpha}(y)+\sin^{\alpha}(y)}\,dy \end{align} where in the last step we have used the substitution $y=\dfrac{\pi}{2}-x$.

Now we can combine the final two steps to get

\begin{align} 2I&=\int_0^{\frac{\pi}{2}}\frac{\cos^{\alpha}(x)}{\cos^{\alpha}(x)+\sin^{\alpha}(x)}\,dx+\int_0^{\frac{\pi}{2}}\frac{\sin^{\alpha}(x)}{\cos^{\alpha}(x)+\sin^{\alpha}(x)}\,dx \\&=\int_0^{\frac{\pi}{2}}\frac{\cos^\alpha(x)+\sin^{\alpha}(x)}{\cos^{\alpha}(x)+\sin^{\alpha}(x)}\,dx \\&=\int_0^{\frac{\pi}{2}} 1\,dx \\&=\frac{\pi}{2} \end{align}

Hence $I=\dfrac{\pi}{4}$.

Peter Woolfitt
  • 21,114