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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class12  >>  Integral Calculus

Integrate : $\int \large\frac {\sin ^6 x +\cos ^6 x }{\sin ^2 x \cos ^2 x }$$dx$

$(a)\;[\tan x -\cot x -3x+c] \\(b)\;[\sin x - cosec x +c] \\(c)\;[\tan x -\sec x -3x +c] \\ (d)\;None$

1 Answer

By using Algebraic formula:-
$A^3+B^3= (a+b)(a^2+b^2-ab)$
=> $ \int \large\frac{(\sin ^2 x)^3 +(\cos ^2 x)^3}{\sin ^2 x \cos ^2 x }$$dx$
=> $\int (\sin ^2 x + \cos ^2 x ) \large\frac{(\sin ^4 x +\cos ^4 x -\sin ^2 x \cos ^2 x )}{\sin ^2 x \cos ^2x }$$dx$
=> $\int 1. \{ \tan ^2 x +\cot ^2 x -1\}$
=> $ (\sec^2 x -1) + (cosec ^2 x -1)-1\}$$dx$
$[\tan x -\cot x -3x+c]$
Hence a is the correct answer.
answered Jan 2, 2014 by meena.p
 
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