Question

will mark brianliest if correct​

Answer 1

Answer:

$- \dfrac{sin2x}{2} \: + \: c$

Step-by-step explanation:

$\bold{To \: find} = > \\ \displaystyle\int \dfrac{ {sin}^{8}x - {cos}^{8}x }{1 - 2 {sin}^{2} {cos}^{2} x } dx$

$\bold{Concept \: used} = > \\1)( {a}^{2} - {b}^{2} ) = (a + b) \: (a - b)$

$2) {sin}^{2} x + {cos}^{2} x = 1$

$3) {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab$

$4) {cos}^{2} x \: - \: {sin}^{2} x = cos2x$

$\bold{Solution} = > \\ \displaystyle\int\dfrac{ {sin}^{8}x - {cos}^{8} x }{1 - 2 {sin}^{2} x {cos}^{2} x} dx$

$=\displaystyle\int \dfrac{ {( {sin}^{4}x )}^{2} - ( {cos}^{4} x) ^{2} }{(1 )^{2} - 2 {sin}^{2}x \: {cos}^{2} x } \: \: dx$

$=\displaystyle\int \dfrac{( {sin}^{4}x + {cos}^{4} x) \: ( {sin}^{4}x - {cos}^{4} x }{ {( {sin}^{2}x + {cos}^{2}x) }^{2} - 2 {sin}^{2} {cos}^{2}x } dx$

$= \displaystyle\int \dfrac{( {sin}^{4}x + {cos}^{4} x) \: ( {sin}^{4}x - {cos}^{4}x) }{( {sin}^{4} x + {cos}^{4} x + 2 {sin}^{2} x {cos}^{2} x - 2 {sin}^{2}x {cos}^{2}x } \: dx$

$= \displaystyle\int \dfrac{( {sin}^{4}x + {cos}^{4} x) \: ( {sin}^{4} x - {cos}^{4}x) }{( {sin}^{4}x + {cos}^{4} x)} dx$

$( {sin}^{4} x + {cos}^{4} x) \: cancel \: out \: from \: numerator \: and \: denominator$

$= \displaystyle\int ( {sin}^{4} x - {cos}^{4} ) \: dx$

$= \displaystyle\int( {sin}^{2} x + {cos}^{2} x) \: ( {sin}^{2} x - {cos}^{2} x)dx$

$= - \displaystyle\int (1) \: ( {cos}^{2} x - {sin}^{2} x) \: dx$

$= - \displaystyle\int \: cos2x \: dx$

$= - \dfrac{sin2x}{2} + c$