For the integral over [0, π], we have: π an (1) sin(x) x cos (nx)dx (using symmetry of integrand) = -sin (¹ sin 2 sin (x) × cos (nx) dx 2 ( ²17 ) by parts) J në 2 2 n në 2 (integrating
For the integral over [0, π], we have: π an (1) sin(x) x cos (nx)dx (using symmetry of integrand) = -sin (¹ sin 2 sin (x) × cos (nx) dx 2 ( ²17 ) by parts) J në 2 2 n në 2 (integrating
For the integral over [0, π], we have: π an (1) sin(x) x cos (nx)dx (using symmetry of integrand) = -sin (¹ sin 2 sin (x) × cos (nx) dx 2 ( ²17 ) by parts) J në 2 2 n në 2 (integrating
How we doing integration by part to obtain the last equation? Explain with details
Transcribed Image Text:For the integral over [0, π], we have:
π
an
(7) sin(x) = cos (nx)dx
2
2
(²) f sin (x) x cos (nx) dx
(using symmetry of integrand) =
sin (-)-sin (7)
2 2
2
n
(²/1)
by parts)
(integrating
With differentiation, one of the major concepts of calculus. Integration involves the calculation of an integral, which is useful to find many quantities such as areas, volumes, and displacement.
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![For the integral over [0, π], we have:
π
an
(7) sin(x) = cos (nx)dx
2
2
(²) f sin (x) x cos (nx) dx
(using symmetry of integrand) =
sin (-)-sin (7)
2 2
2
n
(²/1)
by parts)
(integrating](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd3b7da89-480b-4b70-a8d0-d00beacb0479%2Fa8235ccc-a9e9-455c-a692-e028d603e1c4%2Fap823bd_processed.jpeg&w=3840&q=75)
