Trigonometry (5)

Calculus Level 4

$\large \int_0^{\pi /2} \dfrac1{1 + (\cot x)^{5/9} } \, dx$

Compute the integral above. Give your answer to 3 decimal places.

For your final step, use the approximation $\pi = \dfrac{22}7$ .

The answer is 0.78571.

1 solution

Samara Simha Reddy
Apr 23, 2016

$\begin{aligned} \large \displaystyle I = \int_0^{\pi/2} \frac{1}{1 + (\cot x)^{5/9}} . dx = \int_0^{\pi/2} \frac{(\sin x)^{5/9}}{(\sin x)^{5/9} + (\cos x)^{5/9}}.dx\\ \large \displaystyle = \int_0^{\pi/2} \frac{(\cos x)^{5/9}}{(\sin x)^{5/9} + (\cos x)^{5/9}} .dx \end{aligned}.$

$\large \displaystyle \text{By Adding the above two integrals, We get }\\ \large \displaystyle 2I = \int_0^{\pi/2} \frac{(\cos x)^{5/9} + (\sin x)^{5/9}}{(\sin x)^{5/9} + (\cos x)^{5/9}} . dx = \int_0^{\pi/2} 1.dx = \left[x \right]_0^{\pi/2} = \frac{\pi}{2}$

$\large \displaystyle 2I = \frac{\pi}{2}\\ \large \displaystyle \implies I = \frac{\pi}{4} \approx \color{#D61F06}{\boxed{0.78571}}.$

Hi! Can you please explain how those two integrals are equal? Thank You!

Ashish Sacheti - 5 years, 1 month ago

It is a property of integrals that $\int_a^bf(x)dx$ = $\int_a^bf(a+b-x)dx$ . Here we use the fact that $cosx=sin(\pi/2-x)$ and the integral property above to prove the equality of the two integrals.

Kyle Coughlin - 5 years, 1 month ago

Trigonometry (5)

The answer is 0.78571.

1 solution

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