What if without using Euler's formula ?

Calculus Level 5

∫ 16 sin ⁡ x + 17 cos ⁡ x 18 sin ⁡ x + 19 cos ⁡ x d x = a x + b ln ⁡ ( 18 sin ⁡ x + 19 cos ⁡ x ) c + C \large \int\frac{16\sin{x}+17\cos{x}}{18\sin{x}+19\cos{x}}dx = \frac{ax+b\ln{(18\sin{x}+19\cos{x})}}{c} + C

If a a , b b and c c are co-prime positive integers, find a + b + c a+b+c .


The answer is 1298.

Tommy Li by Tommy Li , 22, Hong Kong

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

Patrick Corn
Mar 20, 2018

Well, here's a roundabout way of doing it.

Note that 16 sin ⁡ x + 17 cos ⁡ x = 2 685 ( − 19 sin ⁡ x + 18 cos ⁡ x ) + 611 685 ( 18 sin ⁡ x + 19 cos ⁡ x ) . 16 \sin x + 17 \cos x = \frac2{685}(-19\sin x + 18\cos x) + \frac{611}{685}(18 \sin x + 19 \cos x). Then the integral equals 2 685 ∫ − 19 sin ⁡ x + 18 cos ⁡ x 18 sin ⁡ x + 19 cos ⁡ x d x + 611 685 ∫ 18 sin ⁡ x + 19 cos ⁡ x 18 sin ⁡ x + 19 cos ⁡ x d x = 2 685 ln ⁡ ( 18 sin ⁡ x + 19 cos ⁡ x ) + 611 685 x + C , \frac2{685}\int \frac{-19\sin x + 18 \cos x}{18 \sin x + 19\cos x} dx + \frac{611}{685} \int \frac{18 \sin x + 19 \cos x}{18 \sin x + 19 \cos x} dx = \frac2{685} \ln(18\sin x + 19\cos x) + \frac{611}{685} x + C, so a = 611 , b = 2 , c = 685 a = 611, b = 2, c = 685 and the answer is 1298 . \fbox{1298}.

By the way, it seems like we're ignoring the problem that the denominator might be zero or negative here.

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Ashutosh Sharma
Mar 19, 2018

https://www.youtube.com/watch?v=2AbBllU4ZE0&index=2&t=0s&list=WL

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...