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Trig Integration (1 Viewer)

Lukybear

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Find the derivative of f(x) = cos x + x sin x. Hence evaluate the definite integral

 

shaon0

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Find the derivative of f(x) = cos x + x sin x. Hence evaluate the definite integral

f'(x)=-sin(x)+sin(x)+xcos(x)
=xcos(x)
Hence,
S {x=0 to x=pi/2} xcos(x) dx = S {x=0 to x=pi/2} d/dx(cos x + x sin x) dx
= {x=0 to x=pi/2} [(cos x + x sin x)]
= pi/2-1
 

Lukybear

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How did you integrate the cosx+xsinx?

O crud, nvm
 

Lukybear

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Also

Derive f(x) = sin2x -2x.cos2x, and us this to find
S{pi/4,0} xsin2x dx
 

life92

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Derive f(x) = sin2x -2x.cos2x, and us this to find
S{pi/4,0} xsin2x dx

f'(x) = 2cos2x - 2 (cos2x -2xsin2x)
= 4xsin2x

therefore,

S{pi/4,0} xsin2x dx = 1/4 [ sin2x - 2xcos2x ] {pi/4,0}
= 1/4 [ (1 - 0) - (0 - 0)
= 1/4
 

shaon0

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Derive f(x) = sin2x -2x.cos2x, and us this to find
S{pi/4,0} xsin2x dx
f'(x)=2cos(2x)-2cos2x+4x.sin2x
=4xsin(2x)

S{pi/4,0} xsin2x dx=(1/4)S{pi/4,0}4xsin2x dx
=(1/4) S{pi/4,0} d/dx(sin2x -2x.cos2x) dx
=(1/4) {pi/4,0}sin2x -2x.cos2x
=(1/4)
 

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