How do u integrate (cos^2 y)(ln t)?

Question

can u list step by step for me..thank-you

the problem is dy/dt = (cos^2 y) ln t

can anyone show me step by step how to solve this DE?

Answer 1

dy/dt = (cos²y)*ln(t)
dy/(cos²y) = ln(t)dt
∫sec²y dy = ∫ln(t)dt
tan(y) = t.ln(t) - t + c

-------------------
∫ln(t)dt
= ∫ln(t).1dt
u = ln(t)
du = (1/t)dt
dv = dt
v = t

∫udv
= uv - ∫vdu
= tln(t) - ∫t(1/t)dt
= tln(t) - ∫1dt
= tln(t) - t

Answer 2

What is the variable of integration here, y or t?
The function of the other one is constant here.
Update: Now I see what the problem is all about.
Let's take it from the top.
dy/dt = cos²y * ln t
Separate the variables.
dy/ cos² y = ln t dt.
sec² y dy = ln t dt.
Now integrate both sides.
The integral of the left side is tan y.
How do we integrate ln t?
Use integration by parts.
U = ln t dV = dt
dU = dt/t V = t
So we get t ln t -t
So finally we have
tan y = t(ln t -1)
or y = arctan( t(ln t-1))+ C.

Answer 3

I see that what you want to do is solve the equation:

dy/dt = (cos^2 y)(ln t)

Rather than integrating both sides immediately, what you do is rewrite it in an easier form.

(1/cos^2 y) dy = (ln t) dt

Now integrate both sides and solve for y

tan y = xln x - x

y = arctan(xlnx -x) + C

(note: i alreasy knew those integrals from memory so i didn't show any steps to calculate them)