intlimits_0^{frac{pi }{2}} frac{dx}{cos^6 x + | vedclass

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(B) Let $I = \int\limits_0^{\frac{\pi }{2}} \frac{dx}{\cos^6 x + \sin^6 x}$.Using the identity $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$,we have $\cos^6 x +

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$\pi$

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(B) Let $I = \int\limits_0^{\frac{\pi }{2}} \frac{dx}{\cos^6 x + \sin^6 x}$.Using the identity $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$,we have $\cos^6 x + \sin^6 x = (\cos^2 x + \sin^2 x)(\cos^4 x - \cos^2 x \sin^2 x + \sin^4 x) = 1 - 3\sin^2 x \cos^2 x$.So,$I = \int\limits_0^{\frac{\pi }{2}} \frac{dx}{1 - 3\sin^2 x \cos^2 x} = \int\limits_0^{\frac{\pi }{2}} \frac{dx}{1 - \frac{3}{4}\sin^2 2x} = \int\limits_0^{\frac{\pi }{2}} \frac{4 dx}{4 - 3\sin^2 2x}$.Let $2x = t$,then $2 dx = dt$,so $dx = \frac{dt}{2}$. When $x=0, t=0$ and when $x=\frac{\pi}{2}, t=\pi$.$I = \int\limits_0^{\pi} \frac{4 \cdot \frac{dt}{2}}{4 - 3\sin^2 t} = 2 \int\limits_0^{\pi} \frac{dt}{4 - 3\sin^2 t} = 4 \int\limits_0^{\frac{\pi}{2}} \frac{dt}{4 - 3\sin^2 t}$.Divide numerator and denominator by $\cos^2 t$: $I = 4 \int\limits_0^{\frac{\pi}{2}} \frac{\sec^2 t dt}{4\sec^2 t - 3	an^2 t} = 4 \int\limits_0^{\frac{\pi}{2}} \frac{\sec^2 t dt}{4(1+	an^2 t) - 3	an^2 t} = 4 \int\limits_0^{\frac{\pi}{2}} \frac{\sec^2 t dt}{4 + 	an^2 t}$.Let $	an t = u$,then $\sec^2 t dt = du$. When $t=0, u=0$ and when $t 	o \frac{\pi}{2}, u 	o \infty$.$I = 4 \int\limits_0^{\infty} \frac{du}{2^2 + u^2} = 4 \left[ \frac{1}{2} 	an^{-1} \left( \frac{u}{2} ight) ight]_0^{\infty} = 2 \left( \frac{\pi}{2} - 0 ight) = \pi$.

$\int\limits_0^{\frac{\pi }{2}} \frac{dx}{\cos^6 x + \sin^6 x}$ is equal to:

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