Calculus masterpiece

Calculus Level 3

0 2 π e x sin ( x ) cos ( x ) d x \large \int_0^{2\pi} e^x \sin(x) \cos(x) \, dx

If the integral above equals to A e B π C \large \frac{A-e^{B\pi}}C for integers A , B A,B and C C , then find the value of A + B + C A+B+C .


The answer is 8.

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2 solutions

We will integrate by parts twice and a little shuffling of I.

I = 0 2 π e x sin ( x ) cos ( x ) d x = 1 2 0 2 π e x sin ( 2 x ) d x = 1 2 0 2 π sin ( 2 x ) d ( e x ) = e x sin ( 2 x ) 2 0 2 π 0 2 π e x cos ( 2 x ) d ( x ) = 0 0 2 π cos ( 2 x ) d ( e x ) = e x cos ( 2 x ) 0 2 π 2 0 2 π sin ( 2 x ) d ( e x ) = e 2 π 1 + 1 2 2 I 5 I = 1 e 2 π 1 I = 1 e 2 π 5 . I = A e B π C . A + B + C = 1 + 2 + 5 = 8 It may be helpful to know the formula given below. e c x sin b x d x = e c x c 2 + b 2 ( c sin b x b cos b x ) = e c x c 2 + b 2 sin ( b x ϕ ) w h e r e ϕ = cos 1 c c 2 + b 2 e c x cos b x d x = e c x c 2 + b 2 ( c cos b x + b sin b x ) = e c x c 2 + b 2 cos ( b x ϕ ) w h e r e ϕ = cos 1 c c 2 + b 2 \large~I= \int_0^{2\pi} e^x \sin(x) \cos(x) \, dx\\ \large=\frac 1 2 \int_0^{2\pi} e^x \sin(2x) \, dx\\ \large= \frac 1 2\int_0^{2\pi} \sin(2x) \, d(e^x)\\ \large = \dfrac{e^x \sin(2x)} 2{\Large |_0^{2\pi} } ~~~~~~~~-~~~~~~ \int_0^{2\pi} e^x \cos(2x) \, d(x)\\\large =~~~~~~~~0~~~~~~~~~~~~ ~~~~~~~~~~-~~~~~~ \int_0^{2\pi} \cos(2x) \, d(e^x) \\\large =-e^x \cos(2x){\Large |_0^{2\pi} }~~~~~~~ - ~~~~~~~2 \int_0^{2\pi} \sin(2x) \, d(e^x)\\= \large - e^{2\pi}*1+1 ~~~~~~~~~~~~~~-~~~~~ 2 *2I \\ \implies \large 5I = 1- e^{2\pi}*1\\ \therefore I=\dfrac {1 -e^{2\pi}} 5. \\\large I=\dfrac {A-e^{B\pi}} C.\\\large \therefore A+B+C=1+2+5=~~~~~~~~~~~\Large \color{#D61F06}{8 } \\\text{It may be helpful to know the formula given below.}\\ \color{#3D99F6}{ \Large \int e^{cx}\sin bx\; \mathrm{d}x = \frac{e^{cx}}{c^2+b^2}(c\sin bx - b\cos bx) \\\Large = \frac{e^{cx}}{\sqrt{c^2+b^2}} *\sin(bx-\phi)\qquad ~where~\phi\large =\cos^{-1} \frac{c}{\sqrt{c^2+b^2}} } \\ \color{#EC7300}{ \Large \int e^{cx}\cos bx\; \mathrm{d}x = \frac{e^{cx}}{c^2+b^2}(c\cos bx + b\sin bx)\\ \Large = \frac{e^{cx}}{\sqrt{c^2+b^2}}\cos(bx-\phi)\qquad~where~\phi\large =\cos^{-1} \frac{c}{\sqrt{c^2+b^2}} } .

@Shashank Rustagi one of my favourite problems!!!!

Noel Lo - 5 years, 10 months ago

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thanks dude, please follow me for more such problems and also try my problem Novice in hide and seek. https://brilliant.org/problems/cool-integral-3/

Shashank Rustagi - 5 years, 10 months ago

can you prove that formula too for other users to understand. I did the same way

Shashank Rustagi - 5 years, 11 months ago

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Just as above by integrating twice by parts, we get the multiple of I on the right hand side. Note that d e c x / d x = c d ( e c x ) a n d d C o s ( b x ) / d x = b S i n ( b x ) \Large~~~~ de^{cx} /dx=c*d(e^{cx})~~and ~~\\\Large dCos(bx)/dx=-b*Sin(bx)

Niranjan Khanderia - 5 years, 11 months ago

Sir, Can you please tell some more important advanced formulae for integration.

mukesh jha - 5 years, 11 months ago

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http://integral-table.com/downloads/integral-table.pdf .It is from here I got the formula. One after google.com.

Niranjan Khanderia - 5 years, 11 months ago

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Thank You Sir

mukesh jha - 5 years, 11 months ago

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@Mukesh Jha You are welcome.

Niranjan Khanderia - 5 years, 11 months ago
Wissam Akil
Nov 12, 2015

by first solving the indefinite integral : e x s i n ( x ) c o s ( x ) d x \int e^x sin(x)cos(x)dx using the integration by parts rule , recall : u . v d x = u v d x ( u ) ( v d x ) d x \int u.vdx=u\int vdx-\int ({u}')(\int vdx)dx using u = e x u=e^x & v = s i n ( x ) c o s ( x ) v=sin(x)cos(x) e x s i n ( x ) c o s ( x ) d x = e x s i n ( x ) c o s ( x ) d x ( d ( e x ) d x ) ( s i n ( x ) c o s ( x ) d x ) d x \int e^x sin(x)cos(x)dx=e^x \int sin(x)cos(x)dx-\int (\frac{\mathrm{d} (e^x)}{\mathrm{d} x})(\int sin(x)cos(x)dx)dx = 1 2 e x . s i n 2 ( x ) 1 2 e x . s i n 2 ( x ) d x =\frac{1}{2}e^x.sin^2(x)-\int \frac{1}{2}e^x.sin^2(x)dx and by solving the last integral : 1 2 e x sin 2 ( x ) d x = 1 2 ( sin 2 ( x ) e x sin ( 2 x ) e x d x ) \frac{1}{2}\int \:e^x\sin ^2\left(x\right)dx=\frac{1}{2}\left(\sin ^2\left(x\right)e^x-\int \sin \left(2x\right)e^xdx\right) = 1 2 ( e x sin 2 ( x ) ( e x ( 1 2 cos ( 2 x ) ) e x ( 1 2 cos ( 2 x ) ) d x ) ) =\frac{1}{2}\left(e^x\sin ^2\left(x\right)-\left(e^x\left(-\frac{1}{2}\cos \left(2x\right)\right)-\int \:e^x\left(-\frac{1}{2}\cos \left(2x\right)\right)dx\right)\right)

= 1 2 ( e x sin 2 ( x ) ( 1 2 e x cos ( 2 x ) 1 2 e x cos ( 2 x ) d x ) ) =\frac{1}{2}\left(e^x\sin ^2\left(x\right)-\left(-\frac{1}{2}e^x\cos \left(2x\right)-\int \:-\frac{1}{2}e^x\cos \left(2x\right)dx\right)\right) = 1 2 ( e x sin 2 ( x ) ( 1 2 e x cos ( 2 x ) ( 1 2 ( e x 1 2 sin ( 2 x ) e x 1 2 sin ( 2 x ) d x ) ) ) ) =\frac{1}{2}\left(e^x\sin ^2\left(x\right)-\left(-\frac{1}{2}e^x\cos \left(2x\right)-\left(-\frac{1}{2}\left(e^x\frac{1}{2}\sin \left(2x\right)-\int \:e^x\frac{1}{2}\sin \left(2x\right)dx\right)\right)\right)\right) = 1 2 ( e x sin 2 ( x ) ( 1 2 e x cos ( 2 x ) ( 1 2 ( e x 1 2 sin ( 2 x ) 1 2 e x sin ( 2 x ) d x ) ) ) ) =\frac{1}{2}\left(e^x\sin ^2\left(x\right)-\left(-\frac{1}{2}e^x\cos \left(2x\right)-\left(-\frac{1}{2}\left(e^x\frac{1}{2}\sin \left(2x\right)-\frac{1}{2}\int \:e^x\sin \left(2x\right)dx\right)\right)\right)\right) = ( ( 1 5 e x ( 2 cos ( 2 x ) sin ( 2 x ) ) ) + sin 2 ( x ) e x ) 1 2 =\left(-\left(-\frac{1}{5}e^x\left(2\cos \left(2x\right)-\sin \left(2x\right)\right)\right)+\sin ^2\left(x\right)e^x\right)\frac{1}{2} = 1 2 ( e x sin 2 ( x ) + 1 5 e x ( 2 cos ( 2 x ) sin ( 2 x ) ) ) =\frac{1}{2}\left(e^x\sin ^2\left(x\right)+\frac{1}{5}e^x\left(2\cos \left(2x\right)-\sin \left(2x\right)\right)\right) now we know that : e x s i n ( x ) c o s ( x ) d x = 1 2 e x s i n 2 ( x ) 1 2 e x s i n 2 ( x ) 1 10 e x ( 2 c o s ( 2 x ) s i n ( 2 x ) ) \int e^x sin(x)cos(x)dx=\frac{1}{2}e^x sin^2(x)-\frac{1}{2}e^x sin^2(x)-\frac{1}{10}e^x(2cos(2x)-sin(2x)) = 1 10 e x ( 2 c o s ( 2 x ) s i n ( 2 x ) ) =-\frac{1}{10}e^x (2cos(2x)-sin(2x)) now , by applying the limits : ( 1 10 e ( 2 π ) ( 2 c o s ( 4 π ) s i n ( 4 π ) ) ) ( 1 10 e 0 ( 2 c o s ( 0 ) s i n ( 0 ) ) ) (-\frac{1}{10}e^(2\pi) (2cos(4\pi )-sin(4\pi )) )-(-\frac{1}{10}e^0 (2cos(0)-sin(0))) which gives : = 2 10 2 e ( 2 π ) 10 =\frac{2}{10} - \frac{2e^(2\pi)}{10} comparing this with the form of the solution : A C e ( B π ) C = 2 10 2 e ( 2 π ) 10 \frac{A}{C} -\frac{e^(B\pi )}{C}=\frac{2}{10}- \frac{2e^(2\pi)}{10}

which gives : B = 2 , C = 5 , A = 1 B=2,C=5,A=1 that is : A + B + C = 1 + 2 + 5 = 8 A+B+C=1+2+5=8

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