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Calculus Level 4

∫ − π / 2 π / 2 e x cos ⁡ 2 x d x \large \int_{- {\pi}/{2}}^{{\pi}/{2}} e^{x} \cos^2x \, dx Given that the integral above can be expressed as a b sinh ⁡ ( c d π ) \displaystyle \frac{a}{b}\sinh \left( \frac{c}{d} \pi \right) , where a , b , c a,b,c and d d are positive integers with gcd ⁡ ( a , b ) = gcd ⁡ ( c , d ) = 1 \gcd(a,b) = \gcd(c,d) = 1 , find the value of a + b + c + d a+b+c+d .


The answer is 12.

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Mardokay Mosazghi by Mardokay Mosazghi , 22, USA

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1 solution

Chew-Seong Cheong
Feb 14, 2016

∫ − π 2 π 2 e x cos ⁡ 2 x d x = 1 2 ∫ − π 2 π 2 e x ( 1 + cos ⁡ ( 2 x ) ) d x = 1 2 ∫ − π 2 π 2 e x d x + 1 4 ∫ − π 2 π 2 e x ( e 2 x i + e − 2 x i ) d x = 1 2 [ e x ] − π 2 π 2 + 1 4 [ e ( 1 + 2 i ) x 1 + 2 i + e ( 1 − 2 i ) x 1 − 2 i ] − π 2 π 2 = 1 2 ( e π 2 − e − π 2 ) + 1 4 [ e x ( e 2 x i + e − 2 x i − 2 i e 2 x i + 2 i e − 2 x i ) 1 + 4 ] − π 2 π 2 = sinh ⁡ π 2 + 1 2 × 5 [ e x ( cos ⁡ ( 2 x ) + 2 sin ⁡ ( 2 x ) ) ] − π 2 π 2 = sinh ⁡ π 2 + 1 10 [ e π 2 ( cos ⁡ π + 2 sin ⁡ π ) − e − π 2 ( cos ⁡ ( − π ) + 2 sin ⁡ ( − π ) ) ] = sinh ⁡ π 2 + 1 10 ( − e π 2 + e − π 2 ) = sinh ⁡ π 2 − 1 5 sinh ⁡ π 2 = 4 5 sinh ⁡ π 2 \begin{aligned} \int_{-\frac{\pi}{2}}^\frac{\pi}{2} e^x \cos^2 x \space dx & = \frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} e^x \left(1+\cos (2x)\right) \space dx \\ & = \frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} e^x \space dx + \frac{1}{4} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} e^x \left(e^{2xi} + e^{-2xi} \right) \space dx \\ & = \frac{1}{2} \left[e^x\right]_{-\frac{\pi}{2}}^\frac{\pi}{2} + \frac{1}{4} \left[ \frac{e^{(1+2i)x}}{1+2i} + \frac{e^{(1-2i)x}}{1-2i} \right]_{-\frac{\pi}{2}}^\frac{\pi}{2} \\ & = \frac{1}{2} \left(e^\frac{\pi}{2} - e^{-\frac{\pi}{2}} \right) + \frac{1}{4} \left[\frac{e^x\left(e^{2xi} + e^{-2xi} - 2ie^{2xi} + 2ie^{-2xi} \right)}{1+4} \right]_{-\frac{\pi}{2}}^\frac{\pi}{2} \\ & = \sinh \frac{\pi}{2} + \frac{1}{2\times 5} \left[e^x \left(\cos (2x) +2 \sin (2x) \right) \right]_{-\frac{\pi}{2}}^\frac{\pi}{2} \\ & = \sinh \frac{\pi}{2} + \frac{1}{10} \left[e^\frac{\pi}{2} \left(\cos \pi + 2\sin \pi \right) - e^{-\frac{\pi}{2}} \left(\cos (-\pi) + 2\sin (-\pi) \right) \right] \\ & =\sinh \frac{\pi}{2} + \frac{1}{10} \left(-e^\frac{\pi}{2} + e^{-\frac{\pi}{2}} \right) \\ & =\sinh \frac{\pi}{2} - \frac{1}{5}\sinh \frac{\pi}{2} \\ & = \frac{4}{5}\sinh \frac{\pi}{2} \end{aligned}

⇒ a + b + c + d = 4 + 5 + 1 + 2 = 12 \Rightarrow a + b + c + d = 4+5+1+2 = \boxed{12}

6 Helpful 0 Interesting 0 Brilliant 0 Confused

awesome method

Hamza A - 5 years, 4 months ago

lovely exactly as intended(i solved this one by your method)

Mardokay Mosazghi - 5 years, 4 months ago

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Okay. Nice to know that you like it.

Chew-Seong Cheong - 5 years, 4 months ago

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