Counterintuitive integration

Calculus Level 5

π / 2 x 4 cos x + 5 x cos x 4 x 3 sin x 5 sin x x 8 + 10 x 5 + 25 x 2 d x \int_{\pi/2}^{\infty} \dfrac{x^4 \cos x + 5x \cos x - 4x^3 \sin x - 5 \sin x}{x^8 + 10x^5 + 25x^2} \, dx

If the above integral can be represented in the form

1 a ( π 2 ) b + c ( π 2 ) d \dfrac{-1}{a \left( \dfrac{\pi}{2} \right)^b + c \left( \dfrac{\pi}{2} \right)^d}

where b > d b > d , and a , b , c , d a , b, c, d are positive integers, find a + b + c + d a+b+c+d .

Details and assumptions:

  • You can try but neither Mathematica nor Wolfram Alpha will compute this for you.


The answer is 11.

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1 solution

Hobart Pao
Apr 21, 2016

Notice that the integrand can be rewritten as

( cos x ) ( x 4 + 5 x ) ( 4 x 3 + 5 ) ( sin x ) ( x 4 + 5 x ) 2 \dfrac{(\cos x) \left( x^4 + 5x \right) - \left( 4x^3 + 5 \right)(\sin x)}{\left(x^4 + 5x\right)^2}

Doesn't that look like the quotient rule? It IS the quotient rule.

Then we have

sin x x 4 + 5 x π / 2 \left. \dfrac{\sin x }{x^4 + 5x} \right|_{\pi/2}^{\infty}

which gives us

A number between 1 and 1 1 ( π 2 ) 4 + 5 ( π 2 ) = 1 ( π 2 ) 4 + 5 ( π 2 ) \dfrac{\text{A number between } -1 \text{ and } 1}{\infty} - \dfrac{1}{\left( \dfrac{\pi}{2}\right)^4 + 5 \left(\dfrac{\pi}{2} \right)} = - \dfrac{1}{\left( \dfrac{\pi}{2}\right)^4 + 5 \left(\dfrac{\pi}{2} \right)}

and that gives 1 + 4 + 5 + 1 = 11 1 + 4 + 5 + 1 = \boxed{11}

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