∫ 0 2 π e x sin ( x ) cos ( x ) d x
If the integral above equals to C A − e B π for integers A , B and C , then find the value of A + B + C .
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@Shashank Rustagi one of my favourite problems!!!!
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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/
can you prove that formula too for other users to understand. I did the same way
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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 )
Sir, Can you please tell some more important advanced formulae for integration.
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http://integral-table.com/downloads/integral-table.pdf .It is from here I got the formula. One after google.com.
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Thank You Sir
by first solving the indefinite integral : ∫ e x s i n ( x ) c o s ( x ) d x using the integration by parts rule , recall : ∫ u . v d x = u ∫ v d x − ∫ ( u ′ ) ( ∫ v d x ) d x using u = e x & v = s i n ( x ) c o s ( 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 x d ( e x ) ) ( ∫ s i n ( x ) c o s ( x ) d x ) d x = 2 1 e x . s i n 2 ( x ) − ∫ 2 1 e x . s i n 2 ( x ) d x and by solving the last integral : 2 1 ∫ e x sin 2 ( x ) d x = 2 1 ( sin 2 ( x ) e x − ∫ sin ( 2 x ) e x d x ) = 2 1 ( e x sin 2 ( x ) − ( e x ( − 2 1 cos ( 2 x ) ) − ∫ e x ( − 2 1 cos ( 2 x ) ) d x ) )
= 2 1 ( e x sin 2 ( x ) − ( − 2 1 e x cos ( 2 x ) − ∫ − 2 1 e x cos ( 2 x ) d x ) ) = 2 1 ( e x sin 2 ( x ) − ( − 2 1 e x cos ( 2 x ) − ( − 2 1 ( e x 2 1 sin ( 2 x ) − ∫ e x 2 1 sin ( 2 x ) d x ) ) ) ) = 2 1 ( e x sin 2 ( x ) − ( − 2 1 e x cos ( 2 x ) − ( − 2 1 ( e x 2 1 sin ( 2 x ) − 2 1 ∫ e x sin ( 2 x ) d x ) ) ) ) = ( − ( − 5 1 e x ( 2 cos ( 2 x ) − sin ( 2 x ) ) ) + sin 2 ( x ) e x ) 2 1 = 2 1 ( e x sin 2 ( x ) + 5 1 e x ( 2 cos ( 2 x ) − sin ( 2 x ) ) ) now we know that : ∫ e x s i n ( x ) c o s ( x ) d x = 2 1 e x s i n 2 ( x ) − 2 1 e x s i n 2 ( x ) − 1 0 1 e x ( 2 c o s ( 2 x ) − s i n ( 2 x ) ) = − 1 0 1 e x ( 2 c o s ( 2 x ) − s i n ( 2 x ) ) now , by applying the limits : ( − 1 0 1 e ( 2 π ) ( 2 c o s ( 4 π ) − s i n ( 4 π ) ) ) − ( − 1 0 1 e 0 ( 2 c o s ( 0 ) − s i n ( 0 ) ) ) which gives : = 1 0 2 − 1 0 2 e ( 2 π ) comparing this with the form of the solution : C A − C e ( B π ) = 1 0 2 − 1 0 2 e ( 2 π )
which gives : B = 2 , C = 5 , A = 1 that is : A + B + C = 1 + 2 + 5 = 8
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We will integrate by parts twice and a little shuffling of I.
I = ∫ 0 2 π e x sin ( x ) cos ( x ) d x = 2 1 ∫ 0 2 π e x sin ( 2 x ) d x = 2 1 ∫ 0 2 π sin ( 2 x ) d ( e x ) = 2 e x sin ( 2 x ) ∣ 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 = 5 1 − e 2 π . I = C A − e B π . ∴ 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 = c 2 + b 2 e c x ( c sin b x − b cos b x ) = c 2 + b 2 e c x ∗ sin ( b x − ϕ ) w h e r e ϕ = cos − 1 c 2 + b 2 c ∫ e c x cos b x d x = c 2 + b 2 e c x ( c cos b x + b sin b x ) = c 2 + b 2 e c x cos ( b x − ϕ ) w h e r e ϕ = cos − 1 c 2 + b 2 c .