Feb 7, 2012 Thread starter #1 M Math New member Feb 7, 2012 1 $ \displaystyle \int_{0}^{\pi/2} \frac{dx}{1+(tan(x))^{\sqrt2}}$ Could anyone tell me how to start, In a book ,it was given that $\sqrt2$ does not even matter Last edited: Feb 7, 2012
$ \displaystyle \int_{0}^{\pi/2} \frac{dx}{1+(tan(x))^{\sqrt2}}$ Could anyone tell me how to start, In a book ,it was given that $\sqrt2$ does not even matter
Feb 7, 2012 #2 M Markov Member Feb 1, 2012 149 Replace $\sqrt2$ for any power and put $x\mapsto\dfrac\pi2-x.$
Feb 12, 2012 #3 S sbhatnagar Active member Jan 27, 2012 95 \( \displaystyle \begin{align*} I &=\int_{0}^{\pi/2}\frac{1}{1+\tan^{\sqrt{2}}(x)}dx \\ \Rightarrow I &=\int_{0}^{\pi/2}\frac{\cos^{\sqrt{2}}(x)}{\sin^{\sqrt{2}}(x)+ \cos^{\sqrt{2}}(x)}dx \qquad (1) \end{align*} \) Applying Property, \( \displaystyle \begin{align*} I &= \int_{0}^{\pi/2}\frac{\cos^{\sqrt{2}}\left( \frac{\pi}{2}-x \right)}{\sin^{\sqrt{2}}\left( \frac{\pi}{2}-x \right)+ \cos^{\sqrt{2}}\left( \frac{\pi}{2}-x \right)}dx\\ \Rightarrow I &= \int_{0}^{\pi/2}\frac{\sin^{\sqrt{2}}(x)}{\sin^{\sqrt{2}}(x)+ \cos^{\sqrt{2}}(x)}dx \qquad (2) \end{align*} \) Add equations (1) and (2) and see what happens! Last edited: Feb 12, 2012
\( \displaystyle \begin{align*} I &=\int_{0}^{\pi/2}\frac{1}{1+\tan^{\sqrt{2}}(x)}dx \\ \Rightarrow I &=\int_{0}^{\pi/2}\frac{\cos^{\sqrt{2}}(x)}{\sin^{\sqrt{2}}(x)+ \cos^{\sqrt{2}}(x)}dx \qquad (1) \end{align*} \) Applying Property, \( \displaystyle \begin{align*} I &= \int_{0}^{\pi/2}\frac{\cos^{\sqrt{2}}\left( \frac{\pi}{2}-x \right)}{\sin^{\sqrt{2}}\left( \frac{\pi}{2}-x \right)+ \cos^{\sqrt{2}}\left( \frac{\pi}{2}-x \right)}dx\\ \Rightarrow I &= \int_{0}^{\pi/2}\frac{\sin^{\sqrt{2}}(x)}{\sin^{\sqrt{2}}(x)+ \cos^{\sqrt{2}}(x)}dx \qquad (2) \end{align*} \) Add equations (1) and (2) and see what happens!
Feb 12, 2012 #4 Sherlock Member Jan 28, 2012 59 Let $x = \tan^{-1}{t}$, then: $\begin{aligned} I & = \int_{0}^{\pi/2}\frac{1}{1+\tan^{\alpha}{x}}\;{dx} \\& = \int_{0}^{\infty}\frac{1}{(1+t^2)(1+t^{\alpha})}\;{dt} \end{aligned}$ Range it from $[0, 1]$ and $[1, \infty]$, $\displaystyle I = \int_{0}^{1}\frac{1}{(1+t^2)(1+t^{\alpha})}\;{dt}+\int_{1}^{\infty}\frac{1}{(1+t^2)(1+t^{\alpha})}\;{dt}$ Put $t \mapsto \frac{1}{t}$ for the second one, $\displaystyle \begin{aligned}I & = \int_{0}^{1}\frac{1}{(1+t^2)(1+t^{\alpha})}\;{dt}+\int_{0}^{1}\frac{t^{\alpha}}{(1+t^2(1+t^{\alpha})}\;{dt} \\& = \int_{0}^{1}\frac{1+t^{\alpha}}{(1+t^2)(1+t^{ \alpha})}\;{dt} = \int_{0}^{1}\frac{1}{1+t^2}\;{dt} = \frac{\pi}{4}.\end{aligned}$ Last edited: Feb 12, 2012
Let $x = \tan^{-1}{t}$, then: $\begin{aligned} I & = \int_{0}^{\pi/2}\frac{1}{1+\tan^{\alpha}{x}}\;{dx} \\& = \int_{0}^{\infty}\frac{1}{(1+t^2)(1+t^{\alpha})}\;{dt} \end{aligned}$ Range it from $[0, 1]$ and $[1, \infty]$, $\displaystyle I = \int_{0}^{1}\frac{1}{(1+t^2)(1+t^{\alpha})}\;{dt}+\int_{1}^{\infty}\frac{1}{(1+t^2)(1+t^{\alpha})}\;{dt}$ Put $t \mapsto \frac{1}{t}$ for the second one, $\displaystyle \begin{aligned}I & = \int_{0}^{1}\frac{1}{(1+t^2)(1+t^{\alpha})}\;{dt}+\int_{0}^{1}\frac{t^{\alpha}}{(1+t^2(1+t^{\alpha})}\;{dt} \\& = \int_{0}^{1}\frac{1+t^{\alpha}}{(1+t^2)(1+t^{ \alpha})}\;{dt} = \int_{0}^{1}\frac{1}{1+t^2}\;{dt} = \frac{\pi}{4}.\end{aligned}$