00:01
The way i would approach this problem, 3x to the 5th sign of x cubed dx, is using a u substitution and let u equal x cubed, because then du could equal 3x squared dx.
00:20
And really what i can do, now that i'm thinking about this, is i can rewrite this integral as 3x squared times x cubed, right? that's the same thing, because you can add those exponents to get that.
00:38
Sign of u, and maybe i could even solve for dx here and show you that 1 over 3x squared du is equal to dx and kind of point out this 3x squared will cancel out.
00:57
But now that i'm looking at that is u, is u also equals this.
01:00
So i'm looking at the integral of u, sine of u, d u.
01:07
Now, what i might have to do is instead of u for integration by parts, i can use y and v prime...