One Solution Used Identity of $a^3 + b^3 = (a+b)(a^2 + b^2 - ab)$ . But I Wanted To Ask If There Is Any Other Method Of Solving This Question ??

Question

I already know a solution that uses the identity

$a^3 + b^3 = (a+b)(a^2 + b^2 - ab)$.

But I wanted to ask if there is any other method of solving this question ??

$$\int\frac{\sin^6x + \cos^6x}{\sin^2x\cos^2x}\, dx$$

Answer 1

You can write $\frac{\sin(x)^6+\cos(x)^6}{\sin(x)^2\cos(x)^2}=\frac{\sin(x)^6}{\sin(x)^2\cos(x)^2}+\frac{\cos(x)^6}{\sin(x)^2\cos(x)^2}$

Answer 2

you can write the integral as $$I={\displaystyle\int}\dfrac{\sin^4\left(x\right)}{\cos^2\left(x\right)}\,\mathrm{d}x+{\displaystyle\int}\dfrac{\cos^4\left(x\right)}{\sin^2\left(x\right)}\,\mathrm{d}x$$

$$I={\displaystyle\int}\class{steps-node}{\cssId{steps-node-1}{\sec^2\left(x\right)}}\cdot\class{steps-node}{\cssId{steps-node-2}{\dfrac{\tan^4\left(x\right)}{\left(\tan^2\left(x\right)+1\right)^2}}}\,\mathrm{d}x+{\displaystyle\int}\class{steps-node}{\cssId{steps-node-16}{\sec^2\left(x\right)}}\cdot\class{steps-node}{\cssId{steps-node-17}{\dfrac{1}{\tan^2\left(x\right)\left(\tan^2\left(x\right)+1\right)^2}}}\,\mathrm{d}x$$

$$I={\displaystyle\int}\dfrac{u^4}{\left(u^2+1\right)^2}\,\mathrm{d}u+{\displaystyle\int}\dfrac{1}{u^2\left(u^2+1\right)^2}\,\mathrm{d}u$$

$$I={\displaystyle\int}1\,\mathrm{d}u-{\displaystyle\int}\dfrac{2u^2+1}{\left(u^2+1\right)^2}\,\mathrm{d}u-{\displaystyle\int}\dfrac{1}{u^2+1}\,\mathrm{d}u-{\displaystyle\int}\dfrac{1}{\left(u^2+1\right)^2}\,\mathrm{d}u+{\displaystyle\int}\dfrac{1}{u^2}\,\mathrm{d}u$$

$$2u^2+1=\class{steps-node}{\cssId{steps-node-1}{\class{steps-node}{\cssId{steps-node-2}{2}}\left(u^2+1\right)}}-\class{steps-node}{\cssId{steps-node-3}{1}}$$ $$I_1=\int\frac{2(u^2+1)-1}{(u^2+1)^2}du$$ $$I_1={\displaystyle\int}\left(\dfrac{\class{steps-node}{\cssId{steps-node-4}{\class{steps-node}{\cssId{steps-node-5}{2}}\left(u^2+1\right)}}}{\left(u^2+1\right)^2}-\dfrac{\class{steps-node}{\cssId{steps-node-6}{1}}}{\left(u^2+1\right)^2}\right)\mathrm{d}u$$ $$I_1={\displaystyle\int}\left(\dfrac{2}{u^2+1}-\dfrac{1}{\left(u^2+1\right)^2}\right)\mathrm{d}u$$

$$I={\displaystyle\int}1\,\mathrm{d}u-{\displaystyle\int}\left(\dfrac{2}{u^2+1}-\dfrac{1}{\left(u^2+1\right)^2}\right)\mathrm{d}u -{\displaystyle\int}\dfrac{1}{u^2+1}\,\mathrm{d}u-{\displaystyle\int}\dfrac{1}{\left(u^2+1\right)^2}\,\mathrm{d}u+{\displaystyle\int}\dfrac{1}{u^2}\,\mathrm{d}u$$

Answer 3

Hint: Write your Integrand as $${\frac {3\, \left( \cos \left( x \right) \right) ^{4}-3\, \left( \cos \left( x \right) \right) ^{2}+1}{ \left(1- \cos^2 \left( x \right) \right) \left( \cos \left( x \right) \right) ^{2}}} $$

Writing $$(\sin^2(x))^3+(\cos^2(x))^3$$

Answer 4

Take into account the relation \begin{align} \sin^6x+\cos^6x&=(\sin^2x+\cos^2x)^3-3\sin^4x\cos^2x-3\sin^2x\cos^4x=\\ &=1-3\sin^2x\cos^2x(\sin^2x+\cos^2x)=\\ &=(\sin^2x+\cos^2x)-3\sin^2x\cos^2x=\\ \end{align} we have \begin{align} \frac{\sin^6x+\cos^6x}{\sin^2x\cos^2x}&=\frac{(\sin^2x+\cos^2x)-3\sin^2x\cos^2x}{\sin^2x\cos^2x}=\\ &=\frac{1}{\cos^2x}+\frac{1}{\sin^2x}-3 \end{align}