• Best of luck to the class of 2024 for their HSC exams. You got this!
    Let us know your thoughts on the HSC exams here
  • YOU can help the next generation of students in the community!
    Share your trial papers and notes on our Notes & Resources page
MedVision ad

Trig Integration (1 Viewer)

Lukybear

Active Member
Joined
May 6, 2008
Messages
1,466
Gender
Male
HSC
2010
Find the derivative of f(x) = cos x + x sin x. Hence evaluate the definite integral

 

shaon0

...
Joined
Mar 26, 2008
Messages
2,029
Location
Guess
Gender
Male
HSC
2009
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

Active Member
Joined
May 6, 2008
Messages
1,466
Gender
Male
HSC
2010
How did you integrate the cosx+xsinx?

O crud, nvm
 

Lukybear

Active Member
Joined
May 6, 2008
Messages
1,466
Gender
Male
HSC
2010
Also

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

life92

Member
Joined
May 7, 2008
Messages
81
Gender
Male
HSC
2009
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

...
Joined
Mar 26, 2008
Messages
2,029
Location
Guess
Gender
Male
HSC
2009
Also

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)
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top