If I_1 = int sin^6 x , dx and I_2 = int cos^6 | vedclass

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(C) We have $I_1 + I_2 = \int (\sin^6 x + \cos^6 x) \, dx$. Using the identity $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$, let $a = \sin^2 x$ and $b = \cos^2

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$\frac{1}{32}(20x + 3 \sin 4x) + c$

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(C) We have $I_1 + I_2 = \int (\sin^6 x + \cos^6 x) \, dx$. Using the identity $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$, let $a = \sin^2 x$ and $b = \cos^2 x$: $\sin^6 x + \cos^6 x = (\sin^2 x + \cos^2 x)(\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x)$. Since $\sin^2 x + \cos^2 x = 1$, this simplifies to $\sin^4 x + \cos^4 x - \sin^2 x \cos^2 x$. We can write $\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2 \sin^2 x \cos^2 x = 1 - 2 \sin^2 x \cos^2 x$. Thus, $\sin^6 x + \cos^6 x = 1 - 3 \sin^2 x \cos^2 x = 1 - \frac{3}{4}(2 \sin x \cos x)^2 = 1 - \frac{3}{4} \sin^2(2x)$. Using $\sin^2 	heta = \frac{1 - \cos 2	heta}{2}$, we get $1 - \frac{3}{4} \left( \frac{1 - \cos 4x}{2} ight) = 1 - \frac{3}{8} + \frac{3}{8} \cos 4x = \frac{5}{8} + \frac{3}{8} \cos 4x$. Integrating this: $\int (\frac{5}{8} + \frac{3}{8} \cos 4x) \, dx = \frac{5x}{8} + \frac{3 \sin 4x}{32} + c = \frac{1}{32}(20x + 3 \sin 4x) + c$.

If $I_1 = \int \sin^6 x \, dx$ and $I_2 = \int \cos^6 x \, dx$, then $I_1 + I_2 = $

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