Look at that exponent

Calculus Level 3

0 π 2 1 1 + tan 4 25 ( x ) d x \large \int_{0}^{\frac{\pi}{2}} \dfrac{1}{1+\tan^{\frac{4}{25}}(x)} \mathrm{d}x

If the above integral evaluates to a number that can be written in the form a π b \dfrac{a\pi}{b} , where a a and b b are coprime, positive integers, find a + b a+b .


The answer is 5.

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2 solutions

Chew-Seong Cheong
May 13, 2017

Relevant wiki: Integration Tricks

I = 0 π 2 1 1 + tan 4 25 x d x Using the identity: a b f ( x ) d x = a b f ( a + b x ) d x = 1 2 0 π 2 1 1 + tan 4 25 x + 1 1 + tan 4 25 ( π 2 x ) d x = 1 2 0 π 2 1 1 + tan 4 25 x + 1 1 + cot 4 25 x d x = 1 2 0 π 2 1 1 + tan 4 25 x + 1 1 + 1 tan 4 25 x d x = 1 2 0 π 2 1 1 + tan 4 25 x + tan 4 25 x tan 4 25 x + 1 d x = 1 2 0 π 2 1 d x = x 2 0 π 2 = π 4 \begin{aligned} I & = \int_0^\frac \pi 2 \frac 1{1+\tan^{\frac 4{25}}x} dx & \small \color{#3D99F6} \text{Using the identity: }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx \\ & = \frac 12 \int_0^\frac \pi 2 \frac 1{1+\tan^{\frac 4{25}}x} + \frac 1{1+\tan^{\frac 4{25}}\left(\frac \pi 2 - x\right)}dx \\ & = \frac 12 \int_0^\frac \pi 2 \frac 1{1+\tan^{\frac 4{25}}x} + \frac 1{1+\cot^{\frac 4{25}} x}dx \\ & = \frac 12 \int_0^\frac \pi 2 \frac 1{1+\tan^{\frac 4{25}}x} + \frac 1{1+\frac 1{\tan^{\frac 4{25}}x}} dx \\ & = \frac 12 \int_0^\frac \pi 2 \frac 1{1+\tan^{\frac 4{25}}x} + \frac {\tan^{\frac 4{25}}x}{\tan^{\frac 4{25}}x + 1} dx \\ & = \frac 12 \int_0^\frac \pi 2 1\cdot dx = \frac x2 \ \bigg|_0^\frac \pi 2 = \frac \pi 4 \end{aligned}

a + b = 1 + 4 = 5 \implies a + b = 1 + 4 = \boxed{5}

Sharky Kesa
May 13, 2017

Let I I be equal to the integral in question and let u = π 2 x u=\frac{\pi}{2}-x . By Change of Variable, we have I = 0 π 2 1 1 + cot 4 25 ( x ) d x I = 0 π 2 1 1 + tan 4 25 ( x ) d x 2 I = 0 π 2 1 1 + tan 4 25 ( x ) + 1 1 + cot 4 25 ( x ) d x = 0 π 2 ( 1 + cot 4 25 ( x ) ) + ( 1 + tan 4 25 ( x ) ) ( 1 + tan 4 25 ( x ) ) ( 1 + cot 4 25 ( x ) ) d x = 0 π 2 2 + cot 4 25 ( x ) + tan 4 25 ( x ) 2 + cot 4 25 ( x ) + tan 4 25 ( x ) d x = 0 π 2 1 d x = π 2 I = π 4 \begin{aligned} I &= \displaystyle \int_{0}^{\frac{\pi}{2}} \dfrac{1}{1+\cot^{\frac{4}{25}}(x)} \mathrm{d}x\\ I &= \displaystyle \int_{0}^{\frac{\pi}{2}} \dfrac{1}{1+\tan^{\frac{4}{25}}(x)} \mathrm{d}x\\ \implies 2I &= \displaystyle \int_{0}^{\frac{\pi}{2}} \dfrac{1}{1+\tan^{\frac{4}{25}}(x)} + \dfrac{1}{1+\cot^{\frac{4}{25}}(x)} \mathrm{d}x\\ &= \displaystyle \int_{0}^{\frac{\pi}{2}} \dfrac{\left ( 1+\cot^{\frac{4}{25}}(x) \right ) + \left ( 1+\tan^{\frac{4}{25}}(x) \right )}{\left ( 1+\tan^{\frac{4}{25}}(x) \right ) \left ( 1+\cot^{\frac{4}{25}}(x) \right )} \mathrm{d}x\\ &= \displaystyle \int_{0}^{\frac{\pi}{2}} \dfrac{2 + \cot^{\frac{4}{25}}(x) +\tan^{\frac{4}{25}}(x)}{2 + \cot^{\frac{4}{25}}(x) +\tan^{\frac{4}{25}}(x)} \mathrm{d}x\\ &= \displaystyle \int_{0}^{\frac{\pi}{2}} 1 \mathrm{d}x\\ &= \dfrac{\pi}{2}\\ \implies I &= \dfrac{\pi}{4} \end{aligned} Thus, we must have a = 1 a=1 and b = 4 b=4 , so a + b = 5 a+b=5 .

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