- #1

mcelgiraffe

- 10

- 0

I have worked the following integral two different ways and received two different answers.

∫sin^3 xcosx dx+ ∫cosx dx

The first method

= ∫sin^3 xcosx dx+ ∫cosx dx

Let u = sinx

du/dx = cosx

dx = 1/cosx du

= ∫u^3 cosx*(1/cosx) du + ∫cosx dx

= ∫u^3 du + ∫cosx dx

= (1/4) u^4+ sinx+C

= (1/4)sin^4 x+sinx+C

The second method

= ∫sinx(sin^2x)cosx dx + ∫cosxdx

= ∫sinx(1-cos^2x)cosx dx + ∫cosx dx

Let u = cosx

du/dx = -sinx

dx = (-1/sinx) du

= ∫sinx(1-u^2 )u* (-1/sinx) du + ∫cosx dx

= -∫(u-u^3)du+∫cosx dx

= (-1/2)u^2+ (1/4)u^4 + sinx + C

= (-1/2)cos^2x + (1/4)cos^4x + sinx + C

If I graph the two answers they are the same graph, but one is shifted vertically by 1/2. Now as an indefinite integral it seems as if this may only be different values of C. But if this were a definite integral wouldn't we drop the C? Meaning that these two equations are not equal? Have I done something wrong in the second equation?

Thank You, James