Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

$https://latex.codecogs.com/png.image?\dpi{110}\int \frac{\sin^8x - \cos^8x}{1-2.\sin^2x.\cos^2x}\;dx=$	A. sin x + cosx + C
	B. sin x − cos x + C
	C. cos x − sin x + C
	D. sin x . cos x + C
	E. − sin x . cos x + C

[ 6-8655 - op net sinds 15.5.2021-(E)-24.8.2025 ]

Translation in E N G L I S H

$\int \frac{\sin^8x - \cos^8x}{1-2.\sin^2x.\cos^2x}\;dx=$	A. sin x + cosx + C
	B. sin x − cos x + C
	C. cos x − sin x + C
	D. sin x . cos x + C
	E. − sin x . cos x + C

Oplossing - Solution

$\begin{aligned}&\quad\int\,\frac{\sin^8x-\cos^8x}{1-2.\sin^2x.\cos^2x}\:dx\\&=\int\,\frac{(\sin^4x+\cos^4x)(\sin^2x+\cos^2x)(\sin^2x-\cos^2x)}{1-2.\sin^2x.\cos^2x}\:dx\\&=\int\,\frac{(\sin^4x+\cos^4x)(\sin^2x-\cos^2x)}{1-2.\sin^2x.\cos^2x}\:dx\\&=\int\,\frac{\left[(\sin^2x+\cos^2x)-2.\sin^2x.\cos^2x\right].(\sin^2x-\cos^2x)}{2-2.\sin^2x.\cos^2x}\:dx\\&=\int\,(\sin^2x-\cos^2x)\:dx\\&=-\int\,\cos 2x\:dx\\&=-\frac12\int\,\cos2x\:d(2x)\\&=-\frac12\sin 2x+C=-\sin x.\cos x+C\end{aligned}$