Question
Asked Oct 2, 2019
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check_circleExpert Solution
Step 1

Given integral can be written as,

j sin' (π) cos' (πκ) dr = f2sin' ( πι) cos" (πε) cos (π:) &
- 12sin (πx)[cos' (π)] cos (πκ) ά
- 2sin' (πκ) [1- sin' (πι)] cos (πx) &
sin2cos2x
1
help_outline

Image Transcriptionclose

j sin' (π) cos' (πκ) dr = f2sin' ( πι) cos" (πε) cos (π:) & - 12sin (πx)[cos' (π)] cos (πκ) ά - 2sin' (πκ) [1- sin' (πι)] cos (πx) & sin2cos2x 1

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Step 2

Now integrating by subs...

put sin (x t
1
Differentiating with respect to .x, we get
dt
cos(Tx)
dc
dt
cOs (Tx) dx
dt
2sin (x)1-sin' (xx)] cos(zx) di 2/f (1-4.5
_
2
Sf(1-2 +)d
_
(-2 dt
2t
2
T3
t'
C
help_outline

Image Transcriptionclose

put sin (x t 1 Differentiating with respect to .x, we get dt cos(Tx) dc dt cOs (Tx) dx dt 2sin (x)1-sin' (xx)] cos(zx) di 2/f (1-4.5 _ 2 Sf(1-2 +)d _ (-2 dt 2t 2 T3 t' C

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Math

Calculus