Consider the integral ∫3x^2 sin⁡x^3 dx. What substitution should we make to make this integral?

Question

Consider the integral ∫3x² sin⁡x³ dx. What substitution should we make to make this integral?

A. u = 3x²

B. u = sinx

C. u = x²

D. u = x³

∫ e^xdx =e^x /ln e +C

a.

True

b.

False

If we use integration by parts on the integral ∫ x sin x dx, then we should pick u and

dv to be

a.

u = sin x   dv = x dx

b.

u = x   dv = dx

c.

u = x  dv = sin⁡ x

d.

u = x   dv = sin x dx

Answer 1

1 D. u=x^3, du=3x^2 dx

integral of sinudu = -cosu

= -cos(x^3) + C

d(-cos(x^3)+C)/dx = sin(x^3)(3x^2)

2 True

integral of e^x = e^x + C and since ln(e) = 1

integral of e^x = e^x/1 + C = e^x/lne + C

d(e^x+C)/dx = e^x

3 D u=x, dv=sinxdx,

v=-cosx, du = dx

find the integral of xsinx

derivative of a product = 1st factor times derivative of 2nd + 2nd times derivative of 1st

(uv)' = uv' + vu'

integrate both sides

uv = integral of udv + integral of vdu

integral of udv = uv - integral of vdu

integral of xsinxdx = -xcosx -sinx

check the solutioon, take the derivative

d(-xcosx +sinx)/dx

=(-xcosx+sinx)'

=-x(-sinx) +cosx(-1) +cosx

= xsinx -cosx+cosx

=xsinx

Answer 2

1) D

2) True

3) D