Welcome to our community

Be a part of something great, join today!

intergral including tan x

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:

Markov

Member
Feb 1, 2012
149
Replace $\sqrt2$ for any power and put $x\mapsto\dfrac\pi2-x.$
 

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! (Giggle)
 
Last edited:

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: