Question
ज्ञात कीजिए: $\in t \sin^3 x\  dx$

Answer

सर्वसमिका $\sin 3x = 3 \sin x - 4 \sin^3x$ से हम पाते हैं कि
$\sin^3x = \frac{3 \sin x-\sin 3 x}{4}$
इसलिए $\in t \sin^3 x\  dx = \frac{3}{4} \in t \sin \ x\  dx - \frac{1}{4} \in t \sin \ 3 x\  dx$
$=-\frac{3}{4} \cos x +\frac{1}{12} \cos 3 x + C$

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