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Question 118 of 129

Q.∫sin⁡3x dx\displaystyle\int \sin^3 x\, dx is:

(a) −34cos⁡x+cos⁡3x12+c-\dfrac{3}{4}\cos x + \dfrac{\cos 3x}{12} + c
(b) −34cos⁡x−cos⁡3x12+c-\dfrac{3}{4}\cos x - \dfrac{\cos 3x}{12} + c
(c) −34sin⁡x−sin⁡3x12+c-\dfrac{3}{4}\sin x - \dfrac{\sin 3x}{12} + c
(d) 34cos⁡x+cos⁡3x12+c\dfrac{3}{4}\cos x + \dfrac{\cos 3x}{12} + c
Puducherry TnboardTamil Nadu HSC First Year (DGE) Board 2023MCQ· 1mImportance★★★★★
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Using sin⁡3x=3sin⁡x−4sin⁡3x\sin3x=3\sin x-4\sin^3x, we can isolate sin⁡3x\sin^3x and integrate term by term.

From the triple-angle identity, sin⁡3x=3sin⁡x−4sin⁡3x\sin3x = 3\sin x - 4\sin^3x, so:

sin⁡3x=3sin⁡x−sin⁡3x4\sin^3x = \frac{3\sin x - \sin3x}{4}

Integrating:

…

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