int {(sin^4 x - cos^4 x) , dx} = | vedclass

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Question

(B) We are given the integral $I = \int (\sin^4 x - \cos^4 x) \, dx$. Using the algebraic identity $a^2 - b^2 = (a - b)(a + b)$,we can write: $\sin^4

Vedclass · Accepted Answer

$-\frac{\sin 2x}{2} + c$

Vedclass · Answer

(B) We are given the integral $I = \int (\sin^4 x - \cos^4 x) \, dx$. Using the algebraic identity $a^2 - b^2 = (a - b)(a + b)$,we can write: $\sin^4 x - \cos^4 x = (\sin^2 x - \cos^2 x)(\sin^2 x + \cos^2 x)$. Since $\sin^2 x + \cos^2 x = 1$,the expression simplifies to: $\sin^4 x - \cos^4 x = \sin^2 x - \cos^2 x = -(\cos^2 x - \sin^2 x)$. Using the trigonometric identity $\cos 2x = \cos^2 x - \sin^2 x$,we get: $\sin^4 x - \cos^4 x = -\cos 2x$. Now,substitute this into the integral: $I = \int -\cos 2x \, dx = -\frac{\sin 2x}{2} + c$.

$\int {(\sin^4 x - \cos^4 x) \, dx} = $

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