The value of int frac{sin^6 x + cos^6 x}{sin^ | vedclass

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(A) We know that $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$.Let $a = \sin^2 x$ and $b = \cos^2 x$. Then $\sin^6 x + \cos^6 x = (\sin^2 x + \cos^2 x)(\sin^4

Vedclass · Accepted Answer

$	an x - \cot x - 3x + c$, where $c$ is the constant of integration

Vedclass · Answer

(A) We know that $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$.Let $a = \sin^2 x$ and $b = \cos^2 x$. Then $\sin^6 x + \cos^6 x = (\sin^2 x + \cos^2 x)(\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x)$.Since $\sin^2 x + \cos^2 x = 1$, we have $\sin^6 x + \cos^6 x = \sin^4 x - \sin^2 x \cos^2 x + \cos^4 x$.Adding and subtracting $2 \sin^2 x \cos^2 x$, we get $\sin^6 x + \cos^6 x = (\sin^2 x + \cos^2 x)^2 - 3 \sin^2 x \cos^2 x = 1 - 3 \sin^2 x \cos^2 x$.Now, the integral becomes $\int \frac{1 - 3 \sin^2 x \cos^2 x}{\sin^2 x \cos^2 x} dx = \int (\frac{1}{\sin^2 x \cos^2 x} - 3) dx$.Using $1 = (\sin^2 x + \cos^2 x)^2$, we have $\int (\frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x} - 3) dx = \int (\sec^2 x + \csc^2 x - 3) dx$.Integrating term by term, we get $	an x - \cot x - 3x + c$.

The value of $\int \frac{\sin^6 x + \cos^6 x}{\sin^2 x \cos^2 x} dx$ is

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