Question

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

Answer 1

$\int \frac{si{n}^{6}x+co{s}^{6}x}{si{n}^{2}xco{s}^{2}x}dx$

$\int \frac{(si{n}^{2}x{)}^3+(co{s}^{2}x{)}^3}{si{n}^{2}xco{s}^{2}x}dx$

$\left[\right(a+b{)}^3=a^3+b^3+3ab(a+b)\Rightarrow a^2+b^3=(a+b{)}^3-3ab(a+b\left)\right]$

$\int \frac{(si{n}^{2}x+co{s}^{2}x{)}^3-3(si{n}^{2}x\left)\right(co{s}^{2}x\left)\right(si{n}^{2}x+co{s}^{2}x)}{si{n}^{2}xco{s}^{2}x}dx$

$[si{n}^{2}x+co{s}^{2}x=1]$

$\int \frac{(1{)}^3-3(si{n}^{2}x\left)\right(co{s}^{2}x\left)\right(1)}{si{n}^{2}xco{s}^{2}x}dx$

$\int \frac{1-3si{n}^{2}xco{s}^{2}x}{si{n}^{2}xco{s}^{2}x}dx$

$\int \frac{1}{si{n}^{2}xco{s}^{2}x}dx-3\int \frac{si{n}^{2}co{s}^{2}x}{si{n}^{2}xco{s}^{2}x}dx$ $[\int dx=x]$

$\int \frac{si{n}^{2}x+co{s}^{2}x}{si{n}^{2}xco{s}^{2}x}dx-3\int dx$ [putting $1=si{n}^{2}x+co{s}^{2}x$]

$\int \frac{si{n}^{2}x}{si{n}^{2}x}dx+\int \frac{co{s}^{2}x}{si{n}^{2}xco{s}^{2}x}dx-3x$

$\int \frac{1}{co{s}^{2}x}dx+\int \frac{1}{si{n}^{2}x}dx-3x$

$\int se{c}^{2}xdx+\int cos{a}^{2}xdx-3x$ $[\int se{c}^{2}xdx=tanx,\int cose{c}^{2}xdx-cotx]$

$tanx+(-cotx)-3x+c$

$\Rightarrow tanx-cotx-3x+c$