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Integration Of trig. Functions

paulmdrdo

Active member
May 13, 2013
386
please correct me with my solution here.

∫cos72xdx = ∫(cos62x cos2x)dx

=∫(1-sin22x)3cos2x

let u = sin2x;
du = cos2xdx
dx = (du/2cos2x)

∫(1-u2)3cos2x*du/2cos2x
1/2∫(1-u2)3*du
1/2∫(1-u2)2*(1-u2)du
1/2∫[(1-3u2-u4-u6]du
1/2∫du-3/2∫u2du-1/2∫u4du-1/2∫u6du

1/2(u)+1/2(u3)-1/10(u5)-1/14(u7)+C

1/2(sin2x)+1/2(sin32x)-1/10(sin52x)-1/14(sin72x)+C

are they correct?
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,708
please correct me with my solution here.

∫cos72xdx = ∫(cos62x cos2x)dx

=∫(1-sin22x)3cos2x

let u = sin2x;
du = 2cos2xdx
dx = (du/2cos2x)

∫(1-u2)3cos2x*du/2cos2x
1/2∫(1-u2)3*du
1/2∫(1-u2)2*(1-u2)du
1/2∫[(1-3u2-u4-u6]du -u4 should be +3u4
1/2∫du-3/2∫u2du-1/2∫u4du-1/2∫u6du

1/2(u)+1/2(u3)-1/10(u5)-1/14(u7)+C

1/2(sin2x)+1/2(sin32x)-1/10(sin52x)-1/14(sin72x)+C

are they correct?
...
 

paulmdrdo

Active member
May 13, 2013
386
$$(1/2) \sin(2x)-(1/2) \sin^{3}(2x)+(3/10) \sin^{5}(2x)-(1/14) \sin^{7}(2x)+C.$$

Is this the right answer?
 
Last edited by a moderator:

Petrus

Well-known member
Feb 21, 2013
739
1/2(sin2x)-1/2(sin32x)+3/10(sin52x)-1/14(sin72x)+C

Is this the write answer? (I supposed you mean right,correct...)
Yes, well done(Clapping)

Regards,
\(\displaystyle |\pi\rangle\)
 

paulmdrdo

Active member
May 13, 2013
386
Yes, well done(Clapping)

Regards,
\(\displaystyle |\pi\rangle\)
thanks for the correction petrus. my brain is kind of boggled.