Please integrate the following: -4(sinmx)^3dx?

Question

Please integrate -4[sin(mx)]^3dx

thanks :)

Hemant · Accepted Answer

I = ∫ ( - 4 sin³ mx )  dx ...... (1)
...............................................................................

We have

sin ( 3 mx ) = 3 sin mx - 4 sin³ mx

so that

- 4 sin³ mx = sin (3mx) - 3 sin mx.
.........................................................................................................

Hence, from (1),

I = ∫ [ sin (3mx) - 3 sin (mx) ] dx

..= [ - ( cos 3mx ) / (3m) ] - 3 [ - ( cos mx ) / (m) ]  + C

..= ( - 1/ 3m) cos 3mx + ( 3/m ) cos mx + C ...................... Ans.
..........................................................................................

Happy To Help !
..................................................................................

Anonymous · Answer

â« -4(sin mx)Â³ dx
= -4 â« (sin mx)Â³ dx
= -4 â« sin mx sinÂ² mx dx
= -4 â« sin mx (1 - cosÂ² mx) dx
= -4 â« (sin mx - sin mx cosÂ² mx) dx
= -4 â« sin mx dx + 4 â« sin mx cosÂ² mx dx
= 4cos (mx)/m + 4 â« sin mx cosÂ² mx dx + C
Let u = cos mx
du = -msin mx
= 4cos (mx)/m - 4/m â« uÂ²du + C
= 4cos (mx)/m - 4/m (uÂ³/3) + C
= 4cos (mx)/m - 4/m (cosÂ³ mx/3) + C
= 4cos (mx)/m - 4cosÂ³ (mx)/(3m) + C
= 4cos (mx)/m [1 - cosÂ² (mx)/3] + C

Trending News

Please integrate the following: -4(sinmx)^3dx?

2 Answers