This is a trigonometric integration using trig identities. S tanX^3 secX dX S tanX^2 secX tanX dX S (secX^2 -1) secX tanX dX u = secX du = secX tanX S ( u^2 - 1) du 1/3secX^3 - secX + C
-2 plus 5 equals +3
secx is the inverse of cosx. secx=1/cosx. A secant is also a line drawn through the graph that touches two points on a function.
22
8
Infinitely many.
This is a trigonometric integration using trig identities. S tanX^3 secX dX S tanX^2 secX tanX dX S (secX^2 -1) secX tanX dX u = secX du = secX tanX S ( u^2 - 1) du 1/3secX^3 - secX + C
sinx*secx ( secx= 1/cos ) sinx*(1/cosx) sinx/cosx=tanx tanx=tanx
-2 plus 5 equals +3
secx is the inverse of cosx. secx=1/cosx. A secant is also a line drawn through the graph that touches two points on a function.
Sec x dx = sec x (secx + tanx)/ (secx + tanx) dx . therefore the answer is ln |secx + tanx|
XtanX dx formula uv - int v du u = x du = dx dv = tanX dx v = ln(secX) x ln(secX) - int ln(secx) dx = X ln(secx) - x ln(secx) - x + C -----------------------------------------
You can't. You need 2 equations to solve for 2 unknowns
You cannot solve it since it is not a true equation.
divide 20 by 10 which equals 2 the times 2 by 2 which equals 4
2y = 4 (divide both sides by 2 to solve) y = 2
22