2016年12月9日 · The answer is =ln (∣tanx+secx∣)-sinx +C We need, secx=1/cosx cos^2x+sin^2x=1 tanx=sinx/cosx (tanx)'=sec^2x (secx)'=tanx secx intsinxtanxdx=int(sinx*sinxdx)/cosx =intsecxsin^2xdx =intsecx(1-cos^2x)dx =int(secx-cosx)dx=intsecxdx-intcosxdx For the integral of secx, multiply top and bottom by (tanx+secx) intsecxdx=int(secx(tanx+secx)dx)/(tanx +secx) Let u=tanx +secx du=(sec^2x+secxtanx)dx ...

Just do the product rule, followed by the chain rule (on only the ln (u)). Recall that d/ (dx) [cosx] = -sinx, and that d/ (dx) [ln (u (x))] = 1/ (u (x)) (du (x ...

2018年3月29日 · tanx 1 − cotx + cotx 1 −tanx = tanx(− tanx) (1 − cotx)(−tanx) + cotx 1 − tanx = −tan2x 1 − tanx + cotx 1 −tanx [∵ tanx ×cotx = 1] = cotx −tan2x ...

2018年5月10日 · How do you evaluate the integral sec x tan x sec2(x) − sec x dx?

2015年3月27日 · How do you use logarithmic differentiation to find the derivative of y = (tan x) 1 x?

graph{tan x [-10, 10, -5, 5]}. See explanation. graph{tan x [-10, 10, -5, 5]} The number of lines you have to draw depends on the domain provided. If the domain is not specifically provided, you will have to draw them all (that's what the function does). If you're doing it by hand, I would first trace some vertical construction lines (asymptotic lines) at pi/2, -pi/2, etc..., where tan x is ...

2017年8月23日 · Either sinx=sqrt (2/3) and cosx=1/sqrt3 or sinx=-sqrt (2/3) and cosx=-1/sqrt3 As tanx=sqrt2 secx=+-sqrt (1+ (sqrt2)^2)=+-sqrt3 and hence cosx=+-1/sqrt3 sinx=tanxcosx=sqrt2xx (+-1/sqrt3)=+-sqrt (2/3) but as tanx is positive, sinx …

How do you prove tan x + cot x sec x + csc x = 1 cos x + sin x?

2015年9月23日 · Refer to explanation It is tanx+1=0=>tanx=-1=>tanx=tan (-pi/4)=> x=k*pi-pi/4 The general solutions are x=k*pi-pi/4 where k belongs to integers

2016年2月23日 · Perhaps what you mean is to prove that tanx 1 +secx + 1 + secx tanx = 2cscx. To solve this let us start from LHS and prove using other identities that this is equal to RHS.