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

Feb 25, 2018

Therefore

$\int \frac{1}{{\sin}^{4} x + {\cos}^{4} x + {\sin}^{2} x {\cos}^{2} x} \mathrm{dx} = - \frac{1}{6} \ln \left({\tan}^{3} x - 1\right) - \ln \left(\tan x + 1\right) - \frac{2}{\sqrt{3}} {\tan}^{-} 1 \left(\tan x - \frac{1}{2}\right) + C$

#### Explanation:

Hope the question is of the form

$\int \frac{1}{{\sin}^{4} x + {\cos}^{4} x + {\sin}^{2} x {\cos}^{2} x} \mathrm{dx}$

Let
$u = {\cos}^{2} x , v = {\sin}^{2} x$

${u}^{3} - {v}^{3} = \left(u - v\right) \left({u}^{2} + u v + {v}^{2}\right)$

Thus

${u}^{2} + u v + {v}^{2} = \frac{{u}^{3} - {v}^{3}}{u - v}$

$\frac{1}{{u}^{2} + u v + {v}^{2}} = \frac{u - v}{{u}^{3} - {v}^{3}}$

Substituting

$\frac{1}{{\sin}^{4} x + {\cos}^{4} x + {\sin}^{2} x {\cos}^{2} x} = \frac{{\cos}^{2} x - {\sin}^{2} x}{{\left({\cos}^{2} x\right)}^{3} - {\left({\sin}^{2} x\right)}^{3}}$

Dividing throughout by ${\cos}^{6} x$

$\frac{1}{{\sin}^{4} x + {\cos}^{4} x + {\sin}^{2} x {\cos}^{2} x} = \frac{{\sec}^{4} x - {\sec}^{2} x {\tan}^{2} x}{1 - {\tan}^{6} x}$

$\frac{1}{{\sin}^{4} x + {\cos}^{4} x + {\sin}^{2} x {\cos}^{2} x} = \frac{{\sec}^{2} x \left({\sec}^{2} x - {\tan}^{2} x\right)}{1 - {\tan}^{6} x}$

${\sec}^{2} x - {\tan}^{2} x = 1$

$\frac{{\sec}^{2} x}{1 - {\tan}^{6} x}$

$\text{Now, ...}$

$\int \frac{1}{{\sin}^{4} x + {\cos}^{4} x + {\sin}^{2} x {\cos}^{2} x} \mathrm{dx} = \int \frac{{\sec}^{2} x}{1 - {\tan}^{6} x} \mathrm{dx}$

Thus,

If
$t = \tan x$

$\mathrm{dt} = {\sec}^{2} x \mathrm{dx}$

$\int \frac{{\sec}^{2} x}{1 - {\tan}^{6} x} \mathrm{dx} = \int \frac{\mathrm{dt}}{1 - {t}^{6}} \mathrm{dx}$

$\frac{1}{1 - {t}^{6}} = \frac{A}{1 - {t}^{3}} + \frac{B}{1 + {t}^{3}}$

$1 = A \left(1 + {t}^{3}\right) + B \left(1 - {t}^{3}\right)$

$1 = A + A {t}^{3} + B - B {t}^{3}$

$1 = \left(A + B\right) + \left(A - B\right) {t}^{3}$

Equating the coefficients of like powers of t

$1 = A + B$

$0 = A - B$

IE

$A = B$

$A = \frac{1}{2} , B = \frac{1}{2}$

$\frac{1}{1 - {t}^{6}} = \frac{A}{1 - {t}^{3}} + \frac{B}{1 + {t}^{3}}$

$\frac{1}{1 - {t}^{6}} = \frac{\frac{1}{2}}{1 - {t}^{3}} - \frac{\frac{1}{2}}{1 + {t}^{3}}$

$\frac{1}{1 - {t}^{6}} = \frac{1}{2} \left(\frac{1}{1 - {t}^{3}} - \frac{1}{1 + {t}^{3}}\right)$

Let

${I}_{1} = \int \frac{1}{1 - {t}^{3}} \mathrm{dt}$

${I}_{2} = \int \frac{1}{1 + {t}^{3}} \mathrm{dt}$

$\frac{1}{1 - {t}^{3}} = \frac{1}{\left(1 - t\right) \left(1 + t + {t}^{2}\right)} = \frac{C}{1 - t} + \frac{D t + E}{1 + t + {t}^{2}}$

$\frac{1}{\left(1 - t\right) \left(1 + t + {t}^{2}\right)} = \frac{C}{1 - t} + \frac{D t + E}{1 + t + {t}^{2}}$

$1 = C \left(1 + t + {t}^{2}\right) + \left(D t + E\right) \left(1 - t\right)$

$1 = C + C t + C {t}^{2} + D t + E - D {t}^{2} - E t$

$1 = \left(C + E\right) + \left(C + D - E\right) t + \left(C - D\right) {t}^{2}$

Equating the coefficients of like powers of t

$0 = C + E$

$E = - C$

$0 = C + D - E$

$0 = C + D + C$

$0 = 2 C + D$

$D = - 2 C$

$C - D = 1$

$C + 2 C = 1$

$3 C = 1$

$C = \frac{1}{3}$

$D = - \frac{2}{3}$

$E = - \frac{1}{3}$

$\frac{1}{\left(1 - t\right) \left(1 + t + {t}^{2}\right)} = \frac{C}{1 - t} + \frac{D t + E}{1 + t + {t}^{2}}$

$\frac{1}{\left(1 - t\right) \left(1 + t + {t}^{2}\right)} = \frac{\frac{1}{3}}{1 - t} + \frac{- \frac{2}{3} t - \frac{1}{3}}{1 + t + {t}^{2}}$

$\frac{1}{\left(1 - t\right) \left(1 + t + {t}^{2}\right)} = \frac{1}{3} \left(\frac{1}{1 - t} - \frac{2 t + 1}{1 + t + {t}^{2}}\right)$

$\frac{1}{\left(1 - t\right) \left(1 + t + {t}^{2}\right)} = \frac{1}{3} \left(- \frac{1}{t - 1} - \frac{2 t + 1}{{t}^{2} + t + 1}\right)$

$\frac{1}{\left(1 - t\right) \left(1 + t + {t}^{2}\right)} = - \frac{1}{3} \left(\frac{1}{t - 1} + \frac{2 t + 1}{{t}^{2} + t + 1}\right)$

${I}_{1} = \int \frac{1}{1 + {t}^{3}} \mathrm{dt}$

${I}_{1} = \int - \frac{1}{3} \left(\frac{1}{t - 1} + \frac{2 t + 1}{{t}^{2} + t + 1}\right) \mathrm{dt}$

${I}_{1} = - \int \frac{1}{3} \left(\frac{1}{t - 1} + \frac{2 t + 1}{{t}^{2} + t + 1}\right) \mathrm{dt}$

${I}_{1} = - \frac{1}{3} \left(\ln \left(t - 1\right) + \ln \left({t}^{2} + t + 1\right)\right)$

$- \frac{1}{3} \ln \left(t - 1\right) \left({t}^{2} + t + 1\right)$

${I}_{1} = - \frac{1}{3} \ln \left({t}^{3} - 1\right)$

$\frac{1}{1 + {t}^{3}} = \frac{1}{\left(1 + t\right) \left(1 - t + {t}^{2}\right)} = \frac{C}{1 + t} + \frac{D t + E}{1 - t + {t}^{2}}$

$\frac{1}{\left(1 + t\right) \left(1 - t + {t}^{2}\right)} = \frac{C}{1 + t} + \frac{D t + E}{1 - t + {t}^{2}}$

$1 = C \left(1 - t + {t}^{2}\right) + \left(D t + E\right) \left(1 + t\right)$

$1 = C - C t + C {t}^{2} + D t + E + D {t}^{2} + E t$

$1 = \left(C + E\right) + \left(- C + D - E\right) t + \left(C + D\right) {t}^{2}$

Equating the coefficients of like powers of t

$0 = C + E$

$E = - C$

$0 = - C + D - E$

$0 = D$

$D = 0$

$C + D = 1$

$C + 0 = 1$

$C = 1$

$D = - \frac{2}{3}$

$E = - 1$

$\frac{1}{\left(1 + t\right) \left(1 - t + {t}^{2}\right)} = \frac{1}{1 + t} + \frac{0 t - 1}{1 - t + {t}^{2}}$

$\frac{1}{\left(1 + t\right) \left(1 - t + {t}^{2}\right)} = \frac{1}{1 + t} - \frac{1}{1 - t + {t}^{2}}$

${I}_{2} = \int \frac{1}{1 + {t}^{3}} \mathrm{dt}$

${I}_{2} = \int \left(\frac{1}{1 + t} - \frac{1}{1 - t + {t}^{2}}\right) \mathrm{dt}$

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

$\int \left(\frac{1}{1 - t + {t}^{2}}\right) \mathrm{dt} = \int \left(\frac{1}{{t}^{2} - t + 1}\right) \mathrm{dt}$

Completing the squares

${t}^{2} - t + 1 = {t}^{2} - 2 \cdot \frac{1}{2} \cdot t + {\left(\frac{1}{2}\right)}^{2} + 1 - {\left(\frac{1}{2}\right)}^{2}$

${\left(t - \frac{1}{2}\right)}^{2} + \frac{3}{4}$

${\left(t - \frac{1}{2}\right)}^{2} + {\left(\frac{\sqrt{3}}{2}\right)}^{2}$

Thus,

$\int \left(\frac{1}{{t}^{2} - t + 1}\right) \mathrm{dt} = \int \frac{1}{{\left(t - \frac{1}{2}\right)}^{2} + {\left(\frac{\sqrt{3}}{2}\right)}^{2}} \mathrm{dt}$

$\int \frac{1}{{\left(t - \frac{1}{2}\right)}^{2} + {\left(\frac{\sqrt{3}}{2}\right)}^{2}} \mathrm{dt} = \frac{1}{\frac{\sqrt{3}}{2}} {\tan}^{-} 1 \left(t - \frac{1}{2}\right)$

$\int \frac{1}{{\left(t - \frac{1}{2}\right)}^{2} + {\left(\frac{\sqrt{3}}{2}\right)}^{2}} \mathrm{dt} = \left(\frac{2}{\sqrt{3}}\right) {\tan}^{-} 1 \left(t - \frac{1}{2}\right)$

${I}_{2} = \ln \left(t + 1\right) + \left(\frac{2}{\sqrt{3}}\right) {\tan}^{-} 1 \left(t - \frac{1}{2}\right)$

$t = \tan x$

${I}_{1} = - \frac{1}{3} \ln \left({t}^{3} - 1\right)$
${I}_{2} = \ln \left(t + 1\right) + \left(\frac{2}{\sqrt{3}}\right) {\tan}^{-} 1 \left(t - \frac{1}{2}\right)$

$I = \frac{1}{2} \left({I}_{1} - {I}_{2}\right)$

$I = \frac{1}{2} \left(- \frac{1}{3} \ln \left({t}^{3} - 1\right) - \left(\ln \left(t + 1\right) + \left(\frac{2}{\sqrt{3}}\right) {\tan}^{-} 1 \left(t - \frac{1}{2}\right)\right)\right)$

$I = - \frac{1}{6} \ln \left({t}^{3} - 1\right) - \ln \left(t + 1\right) - \frac{2}{\sqrt{3}} {\tan}^{-} 1 \left(t - \frac{1}{2}\right)$

Thus,

$\int \frac{1}{{\sin}^{4} x + {\cos}^{4} x + {\sin}^{2} x {\cos}^{2} x} \mathrm{dx} = - \frac{1}{6} \ln \left({t}^{3} - 1\right) - \ln \left(t + 1\right) - \frac{2}{\sqrt{3}} {\tan}^{-} 1 \left(t - \frac{1}{2}\right)$

$t = \tan x$

$\int \frac{1}{{\sin}^{4} x + {\cos}^{4} x + {\sin}^{2} x {\cos}^{2} x} \mathrm{dx} = - \frac{1}{6} \ln \left({\tan}^{3} x - 1\right) - \ln \left(\tan x + 1\right) - \frac{2}{\sqrt{3}} {\tan}^{-} 1 \left(\tan x - \frac{1}{2}\right) + C$