Calculus. Thats it.

Calculus Level 4

∫ 0 π 2 sin ⁡ θ + 3 cos ⁡ θ − 3 13 cos ⁡ θ − 5 d θ = a π b − c d log ⁡ ( e f ) \large\int _0^\pi \frac {2\sin \theta + 3\cos \theta - 3}{ 13\cos \theta - 5} d\theta = \frac {a\pi}b - \frac cd \log \left(\frac ef\right) .

If the equation above holds true for positive integers a a , b b , c c , d d , e e , and f f , where a a , b b , d d , e e , and f f are primes (not necessary distinct), find a + b + c + d + e + f a + b + c + d + e + f .


The answer is 38.

Priyanshu Mishra by Priyanshu Mishra , 20, India

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1 solution

I = ∫ 0 π 2 sin ⁡ θ + 3 cos ⁡ θ − 3 13 cos ⁡ θ − 5 d θ = 2 ∫ 0 π sin ⁡ θ 13 cos ⁡ θ − 5 d θ + 3 ∫ 0 π cos ⁡ θ 13 cos ⁡ θ − 5 d θ − 3 ∫ 0 π 1 13 cos ⁡ θ − 5 d θ Let x = 13 cos ⁡ θ − 5 ⟹ d x = − 13 sin ⁡ θ d θ = − 2 13 ∫ 8 − 18 1 x d x + 3 13 ∫ 0 π 13 cos ⁡ θ − 5 + 5 13 cos ⁡ θ − 5 d θ − 3 ∫ 0 π 1 13 cos ⁡ θ − 5 d θ = − 2 13 log ⁡ ∣ x ∣ ∣ 8 − 18 + 3 13 ∫ 0 π d θ − 24 13 ∫ 0 π 1 13 cos ⁡ θ − 5 d θ Let t = tan ⁡ θ 2 ⟹ d t = 1 2 sec ⁡ 2 θ 2 d θ = − 2 13 log ⁡ 9 4 + 3 13 θ ∣ 0 π − 24 13 ∫ 0 ∞ 1 4 − 9 t 2 d t ⟹ cos ⁡ θ = 1 − t 2 1 + t 2 = − 4 13 log ⁡ 3 2 + 3 13 π − 6 13 ∫ 0 ∞ ( 1 2 − 3 t + 1 2 + 3 t ) d t = 3 13 π − 4 13 log ⁡ 3 2 − 6 13 [ − log ⁡ ∣ 2 − 3 t ∣ 3 + log ⁡ ∣ 2 + 3 t ∣ 3 ] 0 ∞ = 3 13 π − 4 13 log ⁡ 3 2 + 2 13 [ log ⁡ ∣ 2 − 3 t ∣ log ⁡ ∣ 2 + 3 t ∣ ] 0 ∞ = 3 13 π − 4 13 log ⁡ 3 2 + 0 \begin{aligned} I & = \int_0^\pi \frac {2\sin \theta + 3\cos \theta - 3}{13\cos \theta -5} d\theta \\ & = {\color{#3D99F6}2 \int_0^\pi \frac {\sin \theta}{13\cos \theta -5} d\theta}+ 3 \int_0^\pi \frac {\cos \theta}{13\cos \theta -5} d\theta - 3\int_0^\pi \frac 1{13\cos \theta -5} d\theta & \small \color{#3D99F6} \text{Let }x = 13\cos \theta - 5 \implies dx = -13 \sin \theta \ d\theta \\ & = {\color{#3D99F6} - \frac 2{13} \int_8^{-18} \frac 1x dx}+ \frac 3{13} \int_0^\pi \frac {13 \cos \theta - 5 + 5}{13\cos \theta -5} d\theta - 3 \int_0^\pi \frac 1{13\cos \theta -5} d\theta \\ & = - \frac 2{13} \log |x| \bigg|_8^{-18}+ \frac 3{13} \int_0^\pi d\theta - \color{#3D99F6} \frac {24}{13} \int_0^\pi \frac 1{13\cos \theta -5} d\theta & \small \color{#3D99F6} \text{Let }t = \tan \frac \theta 2 \implies dt = \frac 12 \sec^2 \frac \theta 2 \ d\theta \\ & = - \frac 2{13} \log \frac 94 + \frac 3{13} \theta \bigg|_0^\pi - \color{#3D99F6} \frac {24}{13} \int_0^\infty \frac 1{4-9t^2} dt & \small \color{#3D99F6} \implies \cos \theta = \frac {1-t^2}{1+t^2} \\ & = - \frac 4{13} \log \frac 32 + \frac 3{13}\pi - \frac 6{13} \int_0^\infty \left(\frac 1{2-3t} + \frac 1{2+3t} \right)dt \\ & = \frac 3{13}\pi - \frac 4{13} \log \frac 32 - \frac 6{13} \left[-\frac {\log |2-3t|}3 + \frac {\log|2+3t|}3 \right]_0^\infty \\ & = \frac 3{13}\pi - \frac 4{13} \log \frac 32 + \frac 2{13} \left[\frac {\log |2-3t|}{\log|2+3t|} \right]_0^\infty \\ & = \frac 3{13}\pi - \frac 4{13} \log \frac 32 + 0 \end{aligned}

⟹ a + b + c + d + e + f = 3 + 13 + 4 + 13 + 3 + 2 = 38 \implies a+b+c+d+e+f = 3+13+4+13+3+2 = \boxed{38} .

2 Helpful 0 Interesting 0 Brilliant 0 Confused

@Priyanshu Mishra , you have to mention that a a , b b , c c , d d , e e and f f are positive integers. Because a b = 3 13 = 1.5 6.5 = 0.3 1.3 = . . . \dfrac ab = \dfrac 3{13} = \dfrac {1.5}{6.5} = \dfrac {0.3}{1.3} = ... , there are infinite solutions.. Most of the times for ratio, you have to mention they are coprime integers bacause 3 13 = 6 26 = 9 39 = . . . \dfrac 3{13} = \dfrac 6{26} = \dfrac 9{39} = ... , again infinite solution.

Chew-Seong Cheong - 3 years, 5 months ago

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OK THANKS sir for altercation..

Priyanshu Mishra - 3 years, 5 months ago

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