I like integrals 4

Calculus Level 4

0 π / 2 d x 4 cos 2 x + 9 sin 2 x = A B π \large \int_0^{\pi /2} \dfrac{dx}{4\cos^2 x + 9 \sin^2 x} = \dfrac AB \pi

The equation above holds true for coprime positive integers. Find A + B A+B .


The answer is 13.

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1 solution

Divide top and bottom of the integrand by 4 cos 2 ( x ) 4\cos^{2}(x) to end up with

1 4 0 π 2 sec 2 ( x ) 1 + ( 3 2 tan ( x ) ) 2 d x \displaystyle\large \dfrac{1}{4}\int_{0}^{\frac{\pi}{2}} \dfrac{\sec^{2}(x)}{1 + (\frac{3}{2}\tan(x))^{2}} dx .

Now let u = 3 2 tan ( x ) u = \frac{3}{2}\tan(x) . Then d u = 3 2 sec 2 ( x ) d x du = \frac{3}{2}\sec^{2}(x) dx , transforming our integral to

1 6 1 1 + u 2 d u = 1 6 tan 1 ( u ) = 1 6 tan 1 ( 3 2 tan ( x ) ) \displaystyle\large \dfrac{1}{6} \int \dfrac{1}{1 + u^{2}} du = \dfrac{1}{6} \tan^{-1}(u) = \dfrac{1}{6}\tan^{-1}\left(\dfrac{3}{2}\tan(x)\right) ,

which we must then evaluate from x = 0 x = 0 to x = π 2 x = \dfrac{\pi}{2} , (the latter approached from the left). This yields the value

1 6 lim x π 2 tan 1 ( 3 2 tan ( x ) ) 1 6 tan 1 ( 3 2 tan ( 0 ) ) = 1 6 × π 2 0 = π 12 \displaystyle \large \dfrac{1}{6} \lim_{x \to \frac{\pi}{2}^{-}} \tan^{-1}\left(\dfrac{3}{2}\tan(x)\right) - \dfrac{1}{6}\tan^{-1}\left(\dfrac{3}{2}\tan(0)\right) = \dfrac{1}{6} \times \dfrac{\pi}{2} - 0 = \dfrac{\pi}{12} ,

giving us A + B = 1 + 12 = 13 A + B = 1 + 12 = \boxed{13} .

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