MHB Is This Integration of Trigonometric Functions Correct?

[paulmdrdo1]

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]

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?

...

 

[paulmdrdo1]

$$(1/2) \sin(2x)-(1/2) \sin^{3}(2x)+(3/10) \sin^{5}(2x)-(1/14) \sin^{7}(2x)+C.$$

Is this the right answer?

 

[Petrus]

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,
$$|\pi\rangle$$

 

[paulmdrdo1]

Yes, well done(Clapping)

Regards,
$$|\pi\rangle$$

thanks for the correction petrus. my brain is kind of boggled.