Cos(pi n) Symolab - bbbbbbb PDF

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4/9/19, 6(21 pm

Solution

π 1 · ∫0 − sin nπx dx + 1 · ∫0 sin nπx dx = 2 − 2 ( −1 ) −π π π π πn π

( )

( )

n

Steps π 0 1 · ∫−π − sin nπx dx + 1 · ∫0 sin nπx dx π π π π

( )

0

n − sin nπx dx = − 1 ( −1 + ( −1) )

∫−π 0

( )

∫−π

( π) − sin ( nπx ) dx π

Take the constant out:

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n

∫a · f ( x ) dx = a · ∫f( x) dx

0 = −∫−π sin nπx dx π

( )

Cancel the common factor: π 0

= −∫−π sin( nx) dx Hide Steps

Apply u − substitution: u = nx Integral Substitution de!nition

∫ f ( g ( x ) ) · g′ ( x ) dx = ∫ f ( u ) du,

u = g ( x)

Substitute: u = nx Show Steps

du = n dx

⇒ du = ndx ⇒ dx = 1 du n

= ∫sin ( u) 1 du n

1 Simplify sin ( u ) : n

sin (u ) n

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Step-by-Step Calculator - Symbolab

4/9/19, 6(21 pm

= ∫ sin ( u ) du n

Adjust integral boundaries:

x = −π x=0

⇒ u = −πn

⇒u=0

0 = ∫−πn sin ( u ) du

n

0 = −∫−πn sin (u ) du n

Take the constant out:

∫a · f ( x ) dx = a · ∫f( x) dx

0

= − 1 · ∫−πn sin ( u) du n Use the common integral:

∫sin (u ) du =

−cos ( u )

= − 1 [ −cos( u) ] 0

−πn

n

Compute the boundaries: b

∫a f ( x ) dx = F( b)

[−cos (u ) ] 0−πn

= −1 + ( −1 ) n

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− F( a) = limx → b− ( F( x) ) − limx → a+ ( F( x) )

limu→ −πn+ ( −cos( u) ) = −( −1) n

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limu → −πn + ( −cos( u) ) Plug in the value u = −πn

= −cos ( −πn ) Simplify − cos ( −πn ) :

− ( −1)

n

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−cos ( −πn) Use the following identity:

cos ( −x ) = cos ( x )

= −cos ( πn ) Use the following identity:

cos ( nπ ) = ( −1 ) n

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= −( −1) = −( −1)

4/9/19, 6(21 pm

n

n

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limu→ 0− ( −cos ( u ) ) = −1 limu→ 0− ( −cos( u) ) Plug in the value u = 0

= −cos ( 0 ) Use the following trivial identity:

cos ( 0 ) = 1

= −1 = −1 − ( − ( −1 )

)

n

Simplify

= −1 + ( −1) n = − 1 ( −1 + ( −1 ) n ) n

(

)

π = 1 − 1 ( −1 + ( −1) n ) + 1 · ∫0 sin nπx dx π π n π π

∫0

( )

Show Steps

sin nπx dx = 1 ( − (−1 ) n + 1 )

( π)

(

n

)

= 1 − 1 ( −1 + ( −1) n ) + 1 · 1 ( − ( −1 ) n + 1 ) π

n

Simplify

(

(

(

π

1 − 1 −1 + ( −1) n n π

(

n 1 − 1 −1 + ( −1 ) n π

n

) ) + 1π · n1 ( −( −1) n + 1) :

2 − 2 ( −1 ) πn

n

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) ) + 1π · n1 ( −( −1) n + 1)

Remove parentheses:

( −a ) = −a

= − 1 · 1 ( −1 + ( −1 ) n ) + 1 · 1 ( − ( −1 ) n + 1 ) π

n

π

n

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Step-by-Step Calculator - Symbolab

4/9/19, 6(21 pm

Show Steps

n n 1 1 · −1 + ( −1 ) = −1 + (−1 ) πn π n

(

)

Show Steps

n

n 1 · 1 − ( −1) + 1 = − (−1 ) + 1 πn π n

(

)

n n = − (−1 ) − 1 + 1 − ( −1 )

πn

πn

Apply rule

a ± b = a± b c c c

n ( n ) = − ( −1 ) − 1 − ( −1 ) + 1

πn

(

)

n n Expand − −1 + (−1 ) − (−1 ) + 1:

2 − 2 ( −1)

n

Show Steps

n = 2 − 2( −1)

πn

n = 2 − 2 (−1 )

πn

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