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!