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 ...

2018年3月18日 · #int\ 1/sqrt(tanx)\ dx=int\ 1/sqrt(tanx)*(2sqrt(tanx))/sec^2x\ du=# #=int\ 2/sec^2x\ du# Since we can't integrate #x# 's with respect to #u# , we use the following identity:

2016年7月19日 · When tanx = 0, x = npi for some integer n. tanx = sinx/cosx = 0 <=> sinx = 0 <=> x = npi, n in ZZ Thus ...

2018年7月6日 · By definition: #d/dx tanx = lim_(h->0) (tan(x+h)-tanx)/h# Using the trigonometric formulas for the sum of two angles:

2018年6月7日 · secx definitions: csc=1/sin tan=sin/cos solve: =cscx * tanx =1/sinx*sinx/cosx =1/cancel(sinx)*cancel(sinx)/cosx =1/cosx =secx

2015年4月16日 · First of all the conditions: since there is the y=tanx function, x!=pi/2+kpi since there is the y=cotx function, x!=kpi.

2016年5月28日 · as follows Using formula tan (A+B)= (tanA +tanB)/(1-tanAtanB) in the expression of LHS LHS=tan(x+pi/4)=(tanx+tan(pi/4))/(1-tanx*tan(pi/4))=(1+tanx)/(1-tanx)=RHS proved

2017年1月25日 · x=2npi or x=2npi+pi/2 secx=1+tanx can be written as 1/cosx=1+sinx/cosx and assuming cosx!=0 this 1=cosx ...

2015年4月8日 · Use the derivative of the natural logarithm and the chain rule: d/(dx) ln (g(x)) = 1/(g(x)) * g'(x) (Which can also be written: d/(dx) ln (u) = 1/u (du)/(dx)) You'll ...

2016年2月15日 · Here is the graph for #tan(x)#: graph{tan(x) [-10, 10, -5, 5]} No matter how large a value you choose for #x# the range #tan(x-pi)# to #tan(x+pi)# will cover all values #(-oo,+oo)#