What is f(x) = int xsinx + 2secxtanx +3cos4x dx if f(pi)=-2 ?

1 Answer
Jan 26, 2017

f(x) = 3/4sin4x + 2secx - xcosx + sinx - pi

Explanation:

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!