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Integration of square root tan x dx?
for solving this ..the first thing to do is substitute tanx=t^2 then x=tan inverse t^2 then solve the integral..
What is the integral sec x?
ln|sec x + tan x| + C.
What is the integral of tan x-3 dx?
In this specific example one would need to use the u substitution method. * Set u to be x - 3 * Derive x - 3 * u = x - 3 * du = dx Now that we have integrated u we can remove the x - 3 and substitute in u and remove the dx and substitute in du. This is what we have after substituting: * (the integrand of) tan(u)du Now integrate tan(u)du * the Integral of tan(u)du is: * sec2(u) Now resubstitute what we set as u. In this case we set x - 3 to u. This will give us our final answer and integral of tan(x-3)dx. * sec2(x - 3)
What is the integration of tanx?
The integral of tan(x) dx = ln | sec(x) | + cto solve... tan(x) = sin(x)/cos(x)the integral of (sin(x)/cos(x) dx) ... let u = cos(x) then du = -sin(x) dx= the integral of (1/u -du)= -ln | u | + c= -ln | cos(x) | + c= ln | (cos(x))^-1 | + c ... or ... ln | 1/cos(x) | + c= ln | sec(x) | + c
What is the integral of 1 divided by the cosine of x with respect to x?
∫ 1/cos(x) dx = ln(sec(x) + tan(x)) + C C is the constant of integration.
