Question

Solve y= p tan p+ log (cos p), where, p= dy/dx.​

Answer 1

Given:- y= p tan p+ log (cos p) where, p= dy/dx

Solution:-

y= p tan p+ log (cos p) .......(1)

Differentiating eqn(1) with respect to x we get,

$\frac{dy}{dx} = (tan \: p + p {sec}^{2} \: p) \frac{dp}{dx} +( - tan \: p) \frac{dp}{dx}$

$or \: p= p {sec}^{2}p \frac{dp}{dx}$

[where p=dy/dx]

$or \: {sec}^{2}pdp = dx$

Integrating above eqn we get,

$tan \: p =x + c$

$or \: x = tan \: p - c$

[where c is a arbitrary constant]

Hence, y= p tan p+ log (cos p) and x= tan p - c is the solution of given equation.(Ans)

Answer 2

Given : y = ptanp + log(cosp) where p = dy/dx

To find : x in terms of p

Solution:

y = ptanp + log(cosp)

p = dy/dx

y = ptanp + log(cosp)

differentiating wrt x

=> dy/dx  = p sec²p (dp/dx)  + tanp (dp/dx)  +  (1/Cosp)(-Sinp)(dp/dx)

=> dy/dx  = p sec²p (dp/dx)  + tanp (dp/dx)  - tanp(dp/dx)

=> dy/dx  = p sec²p (dp/dx)

dy/dx = p

=> p = p sec²p (dp/dx)

=> 1 = sec²p (dp/dx)

=> dx = sec²p (dp)

integrating both sides

=> x  = tanp + c

Learn more:

Integration of x²×sec²(x³) dx - Brainly.in

brainly.in/question/18063473

brainly.in/question/19073214

https://brainly.in/question/19151558