int frac{dx}{sin^6 x + cos^6 x} = | vedclass

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Question

(A) We know that $\sin^6 x + \cos^6 x = (\sin^2 x + \cos^2 x)(\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x) = 1 \cdot ((\sin^2 x + \cos^2 x)^2 - 3 \sin^2 x

Vedclass · Accepted Answer

$	an^{-1} (	an x - \cot x) + C$

Vedclass · Answer

(A) We know that $\sin^6 x + \cos^6 x = (\sin^2 x + \cos^2 x)(\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x) = 1 \cdot ((\sin^2 x + \cos^2 x)^2 - 3 \sin^2 x \cos^2 x) = 1 - 3 \sin^2 x \cos^2 x$. Dividing numerator and denominator by $\cos^6 x$,we get: $\int \frac{\sec^6 x dx}{	an^6 x + 1} = \int \frac{\sec^4 x \cdot \sec^2 x dx}{	an^6 x + 1} = \int \frac{(1 + 	an^2 x)^2 \sec^2 x dx}{	an^6 x + 1}$. Let $u = 	an x$,then $du = \sec^2 x dx$. The integral becomes $\int \frac{(1 + u^2)^2}{u^6 + 1} du = \int \frac{u^4 + 2u^2 + 1}{u^6 + 1} du$. Dividing numerator and denominator by $u^3$: $\int \frac{u + \frac{2}{u} + \frac{1}{u^3}}{u^3 + \frac{1}{u^3}} du = \int \frac{u + \frac{1}{u^3} + \frac{2}{u}}{u^3 + \frac{1}{u^3}} du$. Using the substitution $t = 	an x - \cot x = u - \frac{1}{u}$,we have $dt = (1 + \frac{1}{u^2}) du = \frac{u^2 + 1}{u^2} du$. After simplification,the integral evaluates to $	an^{-1}(	an x - \cot x) + C$.

$\int \frac{dx}{\sin^6 x + \cos^6 x} = $

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