Since the derivatives of \\sin(x) and \\cos(x) are cyclical, that is, the fourth derivative of each is again \\sin(x) and \\cos(x), it is easy to determine their integrals by logic. The integral and derivative of \\tan(x) is more complicated, but can be determined by studying the derivative and integral of \\ln(x).

Dec 9, 2016 · 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 ...

Oct 2, 2017 · 158801 views around the world You can reuse this answer ...

Jul 4, 2015 · Using the definition of the integral and the fact that sinx is an odd function, from 0 to 2pi, with equal area under the curve at [0, pi] and above the curve at [pi, 2pi], the integral is 0. This holds true for any time sinx is evaluated with an integral across a domain where it is symmetrically above and below the x-axis. int_0^(2pi) sinxdx = [-cosx]|_(0)^(2pi) = -cos2pi - (-cos0) = -1 - (- 1 ...

Aug 13, 2016 · For the integration of a product of two functions (First function)*(Second function) = f(x), int f(x)*dx = (First function)*int (Second function)*dx - int (d/dx(First function)*int(Second function)*dx. This is called integration by parts. The choice of first function and second function is arbitrary in case of most functions. Here, we have f(x) = …

Feb 25, 2018 · sum_(n=0)^(oo)intsin^n(x)/(n!)dx There's really no way to integrate this. The way to integrate is to think "this is the derivative of what?" Since your original equation is e^sin(x) You can't actually apply this, because it would mean: inte^sin(x)dx=-e^sin(x)/cos(x) This isn't the case, however, because this becomes a quotient rule, which leads to a much more complex function afterwards when ...

Aug 12, 2018 · #I=int(1+sin(x))/(1-sin(x))dx# #=int((1+sin(x))(1+sin(x)))/((1-sin(x))(1+sin(x)))dx# #=int((1+sin(x))^2)/(1-sin(x)^2)dx#

Sep 26, 2017 · We use the property #int_0^af(x)dx=int_0^af(a-x)dx#. hence we can write #I=int_0^(pi/2)logsinxdx=int_0^(pi/2)logsin(pi/2-x)dx#

Jul 26, 2016 · Use integration by parts, which takes the form:. #intudv=uv-intvdu# For #intudv=intx^2sin(x)dx#, we let:. #u=x^2" "=>" "(du)/dx=2x" "=>" "du=2xdx#

Mar 9, 2018 · #I=intx/(1+sinx)dx=int(x(1-sinx))/(1-sin^2x)dx=int(x(1-sinx))/cos^2xdx# #=>I=intxsec^2xdx-intxsecxtanxdx# ...