$$\displaystyle \int \frac {\sin^8 x-\cos^8 x}{(1-2\sin^2x \cos^2x)}dx$$ is equal is to
Given, $$\displaystyle\int { \frac { \sin ^{ 8 }{ x } -\cos ^{ 8 }{ x } }{ 1-2\sin ^{ 2 }{ x\cos ^{ 2 }{ x } } } dx }$$
$$=\displaystyle \int { \dfrac { (\sin ^{ 4 }{ x } -\cos ^{ 4 }{ x)(\sin ^{ 4 }{ x } +\cos ^{ 4 }{ x } ) } }{ 1-2\sin ^{ 2 }{ x\cos ^{ 2 }{ x } } } dx } $$
$$=\displaystyle \int { \dfrac { (\sin ^{ 4 }{ x } -\cos ^{ 4 }{ x)((\sin ^{2}{ x })^2+(\cos ^{2}{ x })^2 +2\sin^2x\cos^2x-2\sin^2x\cos^2x ) } }{ 1-2\sin ^{ 2 }{ x\cos ^{ 2 }{ x } } } dx } $$
$$=\displaystyle \int { \dfrac { (\sin ^{ 4 }{ x } -\cos ^{ 4 }{ x)((\sin ^{2}{ x }+\cos ^{2}{ x })^2 -2\sin^2x\cos^2x) } }{ 1-2\sin ^{ 2 }{ x\cos ^{ 2 }{ x } } } dx } $$
$$\displaystyle=\int { \dfrac { (\sin ^{ 2 }{ x } -\cos ^{ 2 }{ x)(1-2\sin ^{ 2 }{ x\cos ^{ 2 }{ x } } ) } }{ 1-2\sin ^{ 2 }{ x\cos ^{ 2 }{ x } } } dx } $$
$$=\displaystyle\int { (-\cos { 2x) } } dx=-\frac { \sin { 2x } }{ 2 } +C$$
Let $$\alpha \epsilon (0, \pi/2)$$ be fixed. If the integral $$\int \dfrac {\tan x + \tan \alpha}{\tan x - \tan \alpha} dx =$$ $$A (x) \cos 2\alpha + B(x) \sin 2\alpha + C$$, where $$C$$ is a constant of integration, then the functions $$A(x)$$ and $$B(x)$$ are respectively.
$$\int \dfrac {dx}{\cos x + \sqrt {3}\sin x}$$ equals
Find the integral of $$\displaystyle \int \frac {\sec^2x}{co\sec^2x}dx$$
Find the integral of $$\displaystyle \int \frac {2-3 \sin x}{\cos^2x}dx$$
Integrate the function $$\sin (ax + b) \cos (ax + b)$$
Integrate the function $$\displaystyle \frac {2\cos x-3 \sin x}{6 \cos x+4 \sin x}$$
$$\displaystyle \int \frac {dx}{\sin^2 x \cos^2x}$$ equals
$$\int_{\cos{2x}\cos{4x}\cos{6x}dx}$$
Find the integrals of the functions $$\sin x \sin 2x \sin 3x$$
Find the integrals of the functions $$\sin 4x \sin 8x$$