Hope the question is of the form
int1/(sin^4x+cos^4x+sin^2xcos^2x)dx∫1sin4x+cos4x+sin2xcos2xdx
Let
u=cos^2x, v=sin^2xu=cos2x,v=sin2x
u^3-v^3=(u-v)(u^2+uv+v^2)u3−v3=(u−v)(u2+uv+v2)
Thus
u^2+uv+v^2=(u^3-v^3)/(u-v)u2+uv+v2=u3−v3u−v
1/(u^2+uv+v^2)=(u-v)/(u^3-v^3)1u2+uv+v2=u−vu3−v3
Substituting
1/(sin^4x+cos^4x+sin^2xcos^2x)=(cos^2x-sin^2x)/((cos^2x)^3-(sin^2x)^3)1sin4x+cos4x+sin2xcos2x=cos2x−sin2x(cos2x)3−(sin2x)3
Dividing throughout by cos^6xcos6x
1/(sin^4x+cos^4x+sin^2xcos^2x)=(sec^4x-sec^2xtan^2x)/(1-tan^6x)1sin4x+cos4x+sin2xcos2x=sec4x−sec2xtan2x1−tan6x
1/(sin^4x+cos^4x+sin^2xcos^2x)=(sec^2x(sec^2x-tan^2x))/(1-tan^6x)1sin4x+cos4x+sin2xcos2x=sec2x(sec2x−tan2x)1−tan6x
sec^2x-tan^2x=1sec2x−tan2x=1
(sec^2x)/(1-tan^6x)sec2x1−tan6x
"Now, ..."Now, ...
int1/(sin^4x+cos^4x+sin^2xcos^2x)dx=int(sec^2x)/(1-tan^6x)dx∫1sin4x+cos4x+sin2xcos2xdx=∫sec2x1−tan6xdx
Thus,
If
t=tanxt=tanx
dt=sec^2xdxdt=sec2xdx
int(sec^2x)/(1-tan^6x)dx=int(dt)/(1-t^6)dx∫sec2x1−tan6xdx=∫dt1−t6dx
1/(1-t^6)=A/(1-t^3)+B/(1+t^3)11−t6=A1−t3+B1+t3
1=A(1+t^3)+B(1-t^3)1=A(1+t3)+B(1−t3)
1=A+At^3+B-Bt^31=A+At3+B−Bt3
1=(A+B)+(A-B)t^31=(A+B)+(A−B)t3
Equating the coefficients of like powers of t
1=A+B1=A+B
0=A-B0=A−B
IE
A=BA=B
A=1/2, B=1/2A=12,B=12
1/(1-t^6)=A/(1-t^3)+B/(1+t^3)11−t6=A1−t3+B1+t3
1/(1-t^6)=(1/2)/(1-t^3)-(1/2)/(1+t^3)11−t6=121−t3−121+t3
1/(1-t^6)=1/2(1/(1-t^3)-1/(1+t^3))11−t6=12(11−t3−11+t3)
Let
I_1=int1/(1-t^3)dtI1=∫11−t3dt
I_2=int1/(1+t^3)dtI2=∫11+t3dt
1/(1-t^3)=1/((1-t)(1+t+t^2))=C/(1-t)+(Dt+E)/(1+t+t^2)11−t3=1(1−t)(1+t+t2)=C1−t+Dt+E1+t+t2
1/((1-t)(1+t+t^2))=C/(1-t)+(Dt+E)/(1+t+t^2)1(1−t)(1+t+t2)=C1−t+Dt+E1+t+t2
1=C(1+t+t^2)+(Dt+E)(1-t)1=C(1+t+t2)+(Dt+E)(1−t)
1=C+Ct+Ct^2+Dt+E-Dt^2-Et1=C+Ct+Ct2+Dt+E−Dt2−Et
1=(C+E)+(C+D-E)t+(C-D)t^21=(C+E)+(C+D−E)t+(C−D)t2
Equating the coefficients of like powers of t
0=C+E0=C+E
E=-CE=−C
0=C+D-E0=C+D−E
0=C+D+C0=C+D+C
0=2C+D0=2C+D
D=-2CD=−2C
C-D=1C−D=1
C+2C=1C+2C=1
3C=13C=1
C=1/3C=13
D=-2/3D=−23
E=-1/3E=−13
1/((1-t)(1+t+t^2))=C/(1-t)+(Dt+E)/(1+t+t^2)1(1−t)(1+t+t2)=C1−t+Dt+E1+t+t2
1/((1-t)(1+t+t^2))=(1/3)/(1-t)+(-2/3t-1/3)/(1+t+t^2)1(1−t)(1+t+t2)=131−t+−23t−131+t+t2
1/((1-t)(1+t+t^2))=1/3(1/(1-t)-(2t+1)/(1+t+t^2))1(1−t)(1+t+t2)=13(11−t−2t+11+t+t2)
1/((1-t)(1+t+t^2))=1/3(-1/(t-1)-(2t+1)/(t^2+t+1))1(1−t)(1+t+t2)=13(−1t−1−2t+1t2+t+1)
1/((1-t)(1+t+t^2))=-1/3(1/(t-1)+(2t+1)/(t^2+t+1))1(1−t)(1+t+t2)=−13(1t−1+2t+1t2+t+1)
I_1=int1/(1+t^3)dtI1=∫11+t3dt
I_1=int-1/3(1/(t-1)+(2t+1)/(t^2+t+1))dtI1=∫−13(1t−1+2t+1t2+t+1)dt
I_1=-int1/3(1/(t-1)+(2t+1)/(t^2+t+1))dtI1=−∫13(1t−1+2t+1t2+t+1)dt
I_1=-1/3(ln(t-1)+ln(t^2+t+1))I1=−13(ln(t−1)+ln(t2+t+1))
-1/3ln(t-1)(t^2+t+1)−13ln(t−1)(t2+t+1)
I_1=-1/3ln(t^3-1)I1=−13ln(t3−1)
1/(1+t^3)=1/((1+t)(1-t+t^2))=C/(1+t)+(Dt+E)/(1-t+t^2)11+t3=1(1+t)(1−t+t2)=C1+t+Dt+E1−t+t2
1/((1+t)(1-t+t^2))=C/(1+t)+(Dt+E)/(1-t+t^2)1(1+t)(1−t+t2)=C1+t+Dt+E1−t+t2
1=C(1-t+t^2)+(Dt+E)(1+t)1=C(1−t+t2)+(Dt+E)(1+t)
1=C-Ct+Ct^2+Dt+E+Dt^2+Et1=C−Ct+Ct2+Dt+E+Dt2+Et
1=(C+E)+(-C+D-E)t+(C+D)t^21=(C+E)+(−C+D−E)t+(C+D)t2
Equating the coefficients of like powers of t
0=C+E0=C+E
E=-CE=−C
0=-C+D-E0=−C+D−E
0=D0=D
D=0D=0
C+D=1C+D=1
C+0=1C+0=1
C=1C=1
D=-2/3D=−23
E=-1E=−1
1/((1+t)(1-t+t^2))=1/(1+t)+(0t-1)/(1-t+t^2)1(1+t)(1−t+t2)=11+t+0t−11−t+t2
1/((1+t)(1-t+t^2))=1/(1+t)-1/(1-t+t^2)1(1+t)(1−t+t2)=11+t−11−t+t2
I_2=int1/(1+t^3)dtI2=∫11+t3dt
I_2=int(1/(1+t)-1/(1-t+t^2))dtI2=∫(11+t−11−t+t2)dt
int(1/(1+t)dt=ln(t+1)∫(11+tdt=ln(t+1)
int(1/(1-t+t^2))dt=int(1/(t^2-t+1))dt∫(11−t+t2)dt=∫(1t2−t+1)dt
Completing the squares
t^2-t+1=t^2-2*1/2*t+(1/2)^2+1-(1/2)^2t2−t+1=t2−2⋅12⋅t+(12)2+1−(12)2
(t-1/2)^2+3/4(t−12)2+34
(t-1/2)^2+(sqrt3/2)^2(t−12)2+(√32)2
Thus,
int(1/(t^2-t+1))dt=int1/((t-1/2)^2+(sqrt3/2)^2)dt∫(1t2−t+1)dt=∫1(t−12)2+(√32)2dt
int1/((t-1/2)^2+(sqrt3/2)^2)dt=1/(sqrt3/2)tan^-1(t-1/2)∫1(t−12)2+(√32)2dt=1√32tan−1(t−12)
int1/((t-1/2)^2+(sqrt3/2)^2)dt=(2/sqrt3)tan^-1(t-1/2)∫1(t−12)2+(√32)2dt=(2√3)tan−1(t−12)
I_2=ln(t+1)+(2/sqrt3)tan^-1(t-1/2)I2=ln(t+1)+(2√3)tan−1(t−12)
t=tanxt=tanx
I_1=-1/3ln(t^3-1)I1=−13ln(t3−1)
I_2=ln(t+1)+(2/sqrt3)tan^-1(t-1/2)I2=ln(t+1)+(2√3)tan−1(t−12)
I=1/2(I_1-I_2)I=12(I1−I2)
I=1/2(-1/3ln(t^3-1)-(ln(t+1)+(2/sqrt3)tan^-1(t-1/2)))I=12(−13ln(t3−1)−(ln(t+1)+(2√3)tan−1(t−12)))
I=-1/6ln(t^3-1)-ln(t+1)-2/sqrt3tan^-1(t-1/2)I=−16ln(t3−1)−ln(t+1)−2√3tan−1(t−12)
Thus,
int1/(sin^4x+cos^4x+sin^2xcos^2x)dx=-1/6ln(t^3-1)-ln(t+1)-2/sqrt3tan^-1(t-1/2)∫1sin4x+cos4x+sin2xcos2xdx=−16ln(t3−1)−ln(t+1)−2√3tan−1(t−12)
t=tanxt=tanx
int1/(sin^4x+cos^4x+sin^2xcos^2x)dx=-1/6ln(tan^3x-1)-ln(tanx+1)-2/sqrt3tan^-1(tanx-1/2)+C∫1sin4x+cos4x+sin2xcos2xdx=−16ln(tan3x−1)−ln(tanx+1)−2√3tan−1(tanx−12)+C
