#f(x) = intxsinxdx + int2secxtanxdx + int3cos4xdx#
#f(x) = intxsinxdx + 2intsecxtanxdx + 3intcos4xdx#
Let's integrate this in pieces.
The integration of #intxsinxdx#
This is a product, so we will use integration by parts. Let #u = x# and #dv = sinxdx#. Then #du = dx# and #v = -cosx#.
#int(u dv) = uv - int(v du)#
#intxsinxdx = x(-cosx) - int-cosxdx#
#intxsinxdx = -xcosx + intcosxdx#
#intxsinxdx = -xcosx + sinx#
The integration of #2intsecxtanxdx#
The integral #intsecxtanxdx# is known as being equal to #secx#. Hence, #2intsecxtanxdx = 2secx#
The integration of #3intcos4xdx#
This is a substitution problem. Let #u = 4x#. then #du = 4dx# and #dx = (du)/4#.
#3intcos4x= 3int cosu * (du)/4#
#3intcos4x = 3/4intcosudu#
#3intcos4x = 3/4sinu#
#3intcos4x = 3/4sin4x#
Putting it all together...
The function is:
#f(x) = 3/4sin4x + 2secx - xcosx + sinx + C#
And finally...
The last step is to find the value of #C#. We know that when #x= pi#, then #y = -2#.
#-2 = 3/4sin(4pi) + 2sec(pi) - picos(pi) + sin(pi) + C#
#-2 = 0 - 2 + pi + 0 + C#
#C = -pi#
The function therefore is #f(x) = 3/4sin4x + 2secx - xcosx + sinx - pi#.
Hopefully this helps!
