May 6, 2018 · We know that, #color(violet)((1)sin^2theta+cos^2theta=1# #color(blue)((2)1/sintheta=csctheta and costheta/sintheta=cottheta#

This integral is a cyclic one and has to be done two times with the theorem of the integration by parts, that says: intf(x)g'(x)dx=f(x)g(x)-intg(x)f'(x)dx We can assume that f(x)=cosx and g'(x)dx=e^xdx, f'(x)dx=-sinxdx g(x)=e^x, so: I=e^xcosx-inte^x(-sinx)dx= =e^xcosx+inte^xsinxdx=(1). And now...again: f(x)=sinx and g'(x)dx=e^xdx, f'(x)dx=cosxdx …

Jun 24, 2018 · Recall the half-angle identity #cos^2(x/2)=(1+cosx)/2#. We can use this to get back nearer to the original style of the integrand expression. #-sqrt(2)cos(x/2)+C=# #-sqrt(2)sqrt((1+cosx)/2)+C=# #-sqrt(1+cosx)+C#. This is rather unexpectedly beautiful, no? The integral of #sqrt(1-cosx)# is a multiple of #sqrt(1+cosx)#. That doesn't seem at all ...

Mar 6, 2015 · The required integral is I = int[cosx/(sinx)^2 + sinx]dx This can be evaluated as two separate integrals, I = I_1 + I_2 , where I_1 = intcosx/(sinx)^2dx and I_2 = intsinxdx The solution of I_2 is trivial, I_2 = intsinxdx = -cosx + C_2 For I_1: Let sinx = t Differentiating with respect to t, cosxdx/dt = 1 => cosxdx = dt This transforms I_1 to I_1 = int1/t^2dt which has the simple …

Oct 18, 2016 · How do you integrate #tanx / (cosx)^2#? Calculus Introduction to Integration Integrals of Trigonometric Functions. 2 Answers

Mar 15, 2015 · First use a double-angle formula to replace cos(2x) by 2cos^{2}(x)-1. Then distribute cos(x) through to rewrite your integrand as (2cos^{3}(x)-cos(x))sin(x).

Jun 6, 2018 · #int (1/cosx+tanx) dx# The integral of the sum is equal to the sum of the integrals, so we can rewrite the above expression as. #color(springgreen)(int (1/cosx)dx)+color(orchid)(int (tanx)dx)# Integrating #1/cosx# #1/cosx# is equivalent to #secx#. Now, we have the integral. #color(springgreen)(int (secx) dx)#, which evaluates to

#I = int \ cosx \ coshx \ dx# #= int \ cosx \ d(sinh x)# #= cosx sinh x - int \ d( cosx) \ sinh x# #= cosx sinh x + int sin x \ sinh x \ dx#

Apr 15, 2018 · How to integrate (sinx)(cosx)(e^cosx)dx ? Calculus. 2 Answers Yahia M. Apr 15, 2018 ...

Mar 15, 2018 · Rewrite #int((sinx+1/cosx)/(sinx/cosx))dx# #=int(((sinxcosx+1)/cancelcosx)/(sinx/cancelcosx))dx# …