What is f(x) = int xsin2x- sec4x dx if f(pi/12)=-2 ?

1 Answer

color(red)(f(x)=-1/2*x*cos 2x+1/4*sin 2x-1/4*ln(sec 4x+tan 4x)+(pisqrt3)/48-17/8+1/4*ln(2+sqrt3)

Explanation:

Given f(x)=int(x*sin 2x-sec 4x) dx if f(pi/12)=-2

We do the integration separately

integration by parts for int(x sin 2x) dx

Let u=x and dv=sin 2x dx
Let v=-1/2 cos 2x" "and du=dx

integration by parts formula

int u dv=uv-int v du

int(x sin 2x) dx=x*(-1/2 cos 2x)-int (-1/2 cos 2x) dx

int(x sin 2x) dx=-x/2 cos 2x+1/4 sin 2x+C_1

the other integral int sec 4x dx

use the formula
int sec u *du=ln (sec u+tan u)+C_2

int sec 4x dx=1/4int sec 4x* 4*dx=1/4*ln (sec 4x+tan 4x)+C_2

The combination

f(x)=int(x*sin 2x-sec 4x) dx

f(x)=-x/2 cos 2x+1/4 sin 2x-1/4*ln (sec 4x+tan 4x)+C_0

If f(pi/12)=-2 we can solve for C_0

-2=-pi/12*1/2 cos 2(pi/12)+1/4 *sin 2(pi/12)-1/4*ln (sec 4(pi/12)+tan 4(pi/12))+C_0

-2=-pi/24* sqrt3/2+1/4 *1/2-1/4*ln (2+sqrt3)+C_0

-2=(-pi* sqrt3)/48+1/8-1/4*ln (2+sqrt3)+C_0

C_0=(pi* sqrt3)/48-17/8+1/4*ln (2+sqrt3)

Final answer

color(red)(f(x)=-x/2 cos 2x+1/4 sin 2x-1/4*ln (sec 4x+tan 4x)+(pi* sqrt3)/48-17/8+1/4*ln (2+sqrt3))

God bless...I hope the explanation is useful.