1÷sin^4x+cos^4x+sin^2xcos^2x.how will you integrate this?

1 Answer
Feb 25, 2018

Therefore

int1/(sin^4x+cos^4x+sin^2xcos^2x)dx=-1/6ln(tan^3x-1)-ln(tanx+1)-2/sqrt3tan^-1(tanx-1/2)+C

Explanation:

Hope the question is of the form

int1/(sin^4x+cos^4x+sin^2xcos^2x)dx

Let
u=cos^2x, v=sin^2x

u^3-v^3=(u-v)(u^2+uv+v^2)

Thus

u^2+uv+v^2=(u^3-v^3)/(u-v)

1/(u^2+uv+v^2)=(u-v)/(u^3-v^3)

Substituting

1/(sin^4x+cos^4x+sin^2xcos^2x)=(cos^2x-sin^2x)/((cos^2x)^3-(sin^2x)^3)

Dividing throughout by cos^6x

1/(sin^4x+cos^4x+sin^2xcos^2x)=(sec^4x-sec^2xtan^2x)/(1-tan^6x)

1/(sin^4x+cos^4x+sin^2xcos^2x)=(sec^2x(sec^2x-tan^2x))/(1-tan^6x)

sec^2x-tan^2x=1

(sec^2x)/(1-tan^6x)

"Now, ..."

int1/(sin^4x+cos^4x+sin^2xcos^2x)dx=int(sec^2x)/(1-tan^6x)dx

Thus,

If
t=tanx

dt=sec^2xdx

int(sec^2x)/(1-tan^6x)dx=int(dt)/(1-t^6)dx

1/(1-t^6)=A/(1-t^3)+B/(1+t^3)

1=A(1+t^3)+B(1-t^3)

1=A+At^3+B-Bt^3

1=(A+B)+(A-B)t^3

Equating the coefficients of like powers of t

1=A+B

0=A-B

IE

A=B

A=1/2, B=1/2

1/(1-t^6)=A/(1-t^3)+B/(1+t^3)

1/(1-t^6)=(1/2)/(1-t^3)-(1/2)/(1+t^3)

1/(1-t^6)=1/2(1/(1-t^3)-1/(1+t^3))

Let

I_1=int1/(1-t^3)dt

I_2=int1/(1+t^3)dt

1/(1-t^3)=1/((1-t)(1+t+t^2))=C/(1-t)+(Dt+E)/(1+t+t^2)

1/((1-t)(1+t+t^2))=C/(1-t)+(Dt+E)/(1+t+t^2)

1=C(1+t+t^2)+(Dt+E)(1-t)

1=C+Ct+Ct^2+Dt+E-Dt^2-Et

1=(C+E)+(C+D-E)t+(C-D)t^2

Equating the coefficients of like powers of t

0=C+E

E=-C

0=C+D-E

0=C+D+C

0=2C+D

D=-2C

C-D=1

C+2C=1

3C=1

C=1/3

D=-2/3

E=-1/3

1/((1-t)(1+t+t^2))=C/(1-t)+(Dt+E)/(1+t+t^2)

1/((1-t)(1+t+t^2))=(1/3)/(1-t)+(-2/3t-1/3)/(1+t+t^2)

1/((1-t)(1+t+t^2))=1/3(1/(1-t)-(2t+1)/(1+t+t^2))

1/((1-t)(1+t+t^2))=1/3(-1/(t-1)-(2t+1)/(t^2+t+1))

1/((1-t)(1+t+t^2))=-1/3(1/(t-1)+(2t+1)/(t^2+t+1))

I_1=int1/(1+t^3)dt

I_1=int-1/3(1/(t-1)+(2t+1)/(t^2+t+1))dt

I_1=-int1/3(1/(t-1)+(2t+1)/(t^2+t+1))dt

I_1=-1/3(ln(t-1)+ln(t^2+t+1))

-1/3ln(t-1)(t^2+t+1)

I_1=-1/3ln(t^3-1)

1/(1+t^3)=1/((1+t)(1-t+t^2))=C/(1+t)+(Dt+E)/(1-t+t^2)

1/((1+t)(1-t+t^2))=C/(1+t)+(Dt+E)/(1-t+t^2)

1=C(1-t+t^2)+(Dt+E)(1+t)

1=C-Ct+Ct^2+Dt+E+Dt^2+Et

1=(C+E)+(-C+D-E)t+(C+D)t^2

Equating the coefficients of like powers of t

0=C+E

E=-C

0=-C+D-E

0=D

D=0

C+D=1

C+0=1

C=1

D=-2/3

E=-1

1/((1+t)(1-t+t^2))=1/(1+t)+(0t-1)/(1-t+t^2)

1/((1+t)(1-t+t^2))=1/(1+t)-1/(1-t+t^2)

I_2=int1/(1+t^3)dt

I_2=int(1/(1+t)-1/(1-t+t^2))dt

int(1/(1+t)dt=ln(t+1)

int(1/(1-t+t^2))dt=int(1/(t^2-t+1))dt

Completing the squares

t^2-t+1=t^2-2*1/2*t+(1/2)^2+1-(1/2)^2

(t-1/2)^2+3/4

(t-1/2)^2+(sqrt3/2)^2

Thus,

int(1/(t^2-t+1))dt=int1/((t-1/2)^2+(sqrt3/2)^2)dt

int1/((t-1/2)^2+(sqrt3/2)^2)dt=1/(sqrt3/2)tan^-1(t-1/2)

int1/((t-1/2)^2+(sqrt3/2)^2)dt=(2/sqrt3)tan^-1(t-1/2)

I_2=ln(t+1)+(2/sqrt3)tan^-1(t-1/2)

t=tanx

I_1=-1/3ln(t^3-1)
I_2=ln(t+1)+(2/sqrt3)tan^-1(t-1/2)

I=1/2(I_1-I_2)

I=1/2(-1/3ln(t^3-1)-(ln(t+1)+(2/sqrt3)tan^-1(t-1/2)))

I=-1/6ln(t^3-1)-ln(t+1)-2/sqrt3tan^-1(t-1/2)

Thus,

int1/(sin^4x+cos^4x+sin^2xcos^2x)dx=-1/6ln(t^3-1)-ln(t+1)-2/sqrt3tan^-1(t-1/2)

t=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