Given f(x)=int(x*sin 2x-sec 4x) dxf(x)=∫(x⋅sin2x−sec4x)dx if f(pi/12)=-2f(π12)=−2
We do the integration separately
integration by parts for int(x sin 2x) dx∫(xsin2x)dx
Let u=xu=x and dv=sin 2x dxdv=sin2xdx
Let v=-1/2 cos 2x" "v=−12cos2x and du=dxdu=dx
integration by parts formula
int u dv=uv-int v du∫udv=uv−∫vdu
int(x sin 2x) dx=x*(-1/2 cos 2x)-int (-1/2 cos 2x) dx∫(xsin2x)dx=x⋅(−12cos2x)−∫(−12cos2x)dx
int(x sin 2x) dx=-x/2 cos 2x+1/4 sin 2x+C_1∫(xsin2x)dx=−x2cos2x+14sin2x+C1
the other integral int sec 4x dx∫sec4xdx
use the formula
int sec u *du=ln (sec u+tan u)+C_2∫secu⋅du=ln(secu+tanu)+C2
int sec 4x dx=1/4int sec 4x* 4*dx=1/4*ln (sec 4x+tan 4x)+C_2∫sec4xdx=14∫sec4x⋅4⋅dx=14⋅ln(sec4x+tan4x)+C2
The combination
f(x)=int(x*sin 2x-sec 4x) dxf(x)=∫(x⋅sin2x−sec4x)dx
f(x)=-x/2 cos 2x+1/4 sin 2x-1/4*ln (sec 4x+tan 4x)+C_0f(x)=−x2cos2x+14sin2x−14⋅ln(sec4x+tan4x)+C0
If f(pi/12)=-2f(π12)=−2 we can solve for C_0C0
-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=−π12⋅12cos2(π12)+14⋅sin2(π12)−14⋅ln(sec4(π12)+tan4(π12))+C0
-2=-pi/24* sqrt3/2+1/4 *1/2-1/4*ln (2+sqrt3)+C_0−2=−π24⋅√32+14⋅12−14⋅ln(2+√3)+C0
-2=(-pi* sqrt3)/48+1/8-1/4*ln (2+sqrt3)+C_0−2=−π⋅√348+18−14⋅ln(2+√3)+C0
C_0=(pi* sqrt3)/48-17/8+1/4*ln (2+sqrt3)C0=π⋅√348−178+14⋅ln(2+√3)
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))f(x)=−x2cos2x+14sin2x−14⋅ln(sec4x+tan4x)+π⋅√348−178+14⋅ln(2+√3)
God bless...I hope the explanation is useful.
