Question

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

Answer 1

$\int \frac{si{n}^{6}x+co{s}^{6}x}{si{n}^{2}xco{s}^{2}x}dx.=\int \frac{(si{n}^{2}x{)}^3+(co{s}^{2}x{)}^3}{si{n}^{2}x.co{s}^{2}x}dx$

$=\int \frac{(si{n}^{2}x+co{s}^{2}x)(si{n}^{4}x+co{s}^{4}x-si{n}^{2}x.co{s}^{2}x)}{si{n}^{2}x.co{s}^{2}x}dx$

using

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

$=\int \frac{1.1si{n}^{4}x+co{s}^{4}x-si{n}^{2}x.co{s}^{2}x}{si{n}^{2}x.co{s}^{2}x}dx[si{n}^{2}x+co{s}^{2}x=1]$

$=\int [\frac{si{n}^{4}x}{si{n}^{2}x.co{s}^{2}x}+\frac{co{s}^{4}x}{si{n}^{2}xco{s}^{2}x}-\frac{si{n}^{2}xco{s}^{2}x}{si{n}^{2}x.co{s}^{2}x}]dx$

$=\int (ta{n}^{2}x+co{t}^{2}x-1)dx$

$=\int (se{c}^{2}x-1+cose{c}^{2}x-1-1)dx$

$=\int \left[\right(se{c}^{2}x+cose{c}^{2}x)-3]dx$

$=tanx-cotx-3x+c$