Evaluate the following integrals

can anyone help me with these questions i did them my self and i'm not sure if my answers are right
can u please include solutions on how u got ur answer too
thank you

A) lnx/x^2

my answer 2(lnx)^2/2x^3

B) cos^2x sinx dx

my answer sinx - sin^3x/3 + c

∫ lnx/(x^2) dx

Applying integration by parts;
u = lnx
du = 1/x dx
dv = 1/x^2
v = -1/x

uv - ∫ vdu
(lnx)(-1/x) - ∫ (-1/x)(1/x) dx
-lnx/x + ∫ 1/x^2 dx
-lnx/x - 1/x + C (Answer)

Question B:

∫ cos^2x sinx dx

Applying algebraic substitution:
u = cosx
du = -sinx dx
-du = sinx dx

∫ u^2 -du
- ∫ u^2 du
(-1/3)u^3 + C
(-1/3)cos^3x + C (Answer)

Have a good day.

Exploit the formula

integration of (u v ) dx = u integration of (v ) dx - integration of (du/dx integration of (v) dx) dx

A) Integration(ln x/ x^2) dx

Put ln x = u and 1/ x^2 = v

Integration(ln x/ x^2) dx

= ln x integration (1/x^2) dx - integration of (d(ln x)/dx integration of (1/x^2) dx) dx

= ln x (-1)/x - integration (1/x * (-1)/x)dx

= - (ln x)/x + integration (1/x^2) dx
= - ln x / x - 1/x

B) Let cos x = u

=> - sin x dx = du
=> sin x dx = -du

So, integration(cos^2 x sin x) dx
= integration (u^2 (-1) du ) = - integration (u^2) du = - u^3/3 = - cos^3 x/3

Read S as inT cosx.tegral.
Slnx/x^2 dx= = (lnx)Sx^-2dx - S[(1/x)Sx^-2dx]dx = -(lnx)/x - 1/x + c.

Second als is not correct. it should be -(1/3)(cosx)^3+c, just by putting T = cosx

A)
∫ lnx/x^2 dx = - ∫Ln(x) d(1/x) = -(1/x)Ln|x| +∫1/x^2 dx = -(1/x)Ln|x| -1/x+ C

B)
∫ cos^2x sinx dx = -∫ cos^2(x) d(cos (x) = (-1/3)cos^3 (x) + C