tan 6 x \tan^6x

Calculus Level 3

π / 2 π / 2 d x tan 6 x + 1 = ? \large \int_{-\pi/2}^{\pi/2} \frac{dx}{\tan^{6}x+1} = \ ?

2 π 2\pi π 2 \frac{\pi} {2} π 4 \frac{\pi} {4} π \pi 1

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1 solution

Chew-Seong Cheong
Aug 21, 2017

I = π 2 π 2 d x tan 6 x + 1 The integrand is even. = 2 0 π 2 d x tan 6 x + 1 Using a b f ( x ) d x = a b f ( a + b x ) d x = 0 π 2 ( 1 tan 6 x + 1 + 1 tan 6 ( π 2 x ) + 1 ) d x = 0 π 2 ( 1 tan 6 x + 1 + 1 cot 6 x + 1 ) d x = 0 π 2 ( 1 tan 6 x + 1 + tan 6 x 1 + tan 6 x ) d x = 0 π 2 d x = π 2 \begin{aligned} I & = \int_{-\frac \pi 2}^\frac \pi 2 \frac {dx}{\tan^6 x + 1} & \small \color{#3D99F6} \text{The integrand is even.} \\ & = 2 \int_0 ^\frac \pi 2 \frac {dx}{\tan^6 x + 1} & \small \color{#3D99F6} \text{Using }\int_a^b f(x) \ dx = \int_a^b f(a+b-x) \ dx \\ & = \int_0 ^\frac \pi 2 \left( \frac 1{\tan^6 x + 1}+\frac 1{\tan^6 \left(\frac \pi 2 - x \right) + 1} \right) dx \\ & = \int_0 ^\frac \pi 2 \left( \frac 1{\tan^6 x + 1}+\frac 1{\cot^6 x + 1} \right) dx \\ & = \int_0 ^\frac \pi 2 \left( \frac 1{\tan^6 x + 1}+\frac {\tan^6 x}{1+\tan^6 x} \right) dx \\ & = \int_0 ^\frac \pi 2 dx = \boxed{\dfrac \pi 2} \end{aligned}

You really helped me. Thank you

Hussain Alghazal - 3 years, 9 months ago

I still new to these stuffs I need to learn how to use LaTeX.

Hussain Alghazal - 3 years, 9 months ago

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