Question

find integration of sin^6x+sin^6x/sin^2x sin^2x​

Answer 1

$\underline{\texttt{Corrected Question :}}$

$\mathrm{Find\:\int \frac{sin^{6}x+cos^{6}x}{sin^{2}x\:cos^{2}x}\:dx}$

$\underline{\texttt{Solution :}}$

$\mathrm{Now,\:\frac{sin^{6}x+cos^{6}x}{sin^{2}x\:cos^{2}x}}$

$\mathrm{=\frac{(sin^{2}x+cos^{2}x)^{3}-3\:sin^{2}x\:cos^{2}x(sin^{2}x+cos^{2}x)}{sin^{2}x\:cos^{2}x}}$

$\mathrm{=\frac{1-3\:sin^{2}x\:cos^{2}x}{sin^{2}x\:cos^{2}x}}$

$\mathrm{=\frac{1}{sin^{2}x\:cos^{2}x}-3}$

$\mathrm{=cosec^{2}x\:sec^{2}x-3}$

$\mathrm{=(1+cot^{2}x)sec^{2}x-3}$

$\mathrm{=sec^{2}x+cot^{2}x\:sec^{2}x-3}$

$\mathrm{=sec^{2}x+\frac{sec^{2}x}{(tanx)^{2}}-3}$

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

$\mathrm{=\int \{sec^{2}x+\frac{sec^{2}x}{(tanx)^{2}}-3\}\:dx}$

$\mathrm{=\int sec^{2}x\:dx+\int \frac{sec^{2}x}{(tanx)^{2}}dx-3\int dx}$

$\mathrm{=\int sec^{2}x\:dx+\int \frac{d(tanx)}{(tanx)^{2}}-3\int dx}$

$\mathrm{=tanx-\frac{1}{tanx}-3x+C}$

$\texttt{where C is integral constant}$

$\to \boxed{\mathrm{\int \frac{sin^{6}x+cos^{6}x}{sin^{2}x\:cos^{2}x}\:dx=tanx-\frac{1}{tanx}-3x+C}}$

$\texttt{which is the required integral}$

$\underline{\texttt{Rules to remember :}}$

$\mathrm{1.\:\int sec^{2}x\:dx=tanx+C}$

$\texttt{where C is integral constant}$

$\mathrm{2.\:\int x^{n}dx=\frac{x^{n+1}}{n+1}+C}$

$\texttt{where C is integral constant}$

$\mathrm{3.\:\int \frac{d\{f(x)\}}{\{f(x)\}^{2}}=-\frac{1}{f(x)}+C}$

$\texttt{where C is integral constant}$