The value of - sin 3 x-cos 3x dx iswhere C is constant of integration

Let I=∫ (sin ^3 x+cos ^3x) dx Using, sin^3x=14[3sin x sin3x], cos^3x=14[cos3x+3cos x] we get, I=14∫( sin3x+3sin x+cos3x+3cos x)dx =14[cos3x3 3cos x+sin3x3+3sin ...

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Byju's Answer

Standard XII

Mathematics

How to Find the Inverse of a Function

The value of ...

Question

The value of $\int ({sin}^{3} x + {cos}^{3} x) d x$ is
(where $C$ is constant of integration)

\frac{3 sin x + 3 cos x}{12} + \frac{sin 3 x - cos 3 x}{12} + C

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\frac{3 sin x + 3 cos x}{4} + \frac{sin 3 x - cos 3 x}{12} + C

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\frac{sin x - cos x}{4} + \frac{sin 3 x + cos 3 x}{12} + C

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\frac{3 sin x - 3 cos x}{4} + \frac{sin 3 x + cos 3 x}{12} + C

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Solution

The correct option is D $\frac{3 sin x - 3 cos x}{4} + \frac{sin 3 x + cos 3 x}{12} + C$
Let
$I = \int ({sin}^{3} x + {cos}^{3} x) d x$
Using,
${sin}^{3} x = \frac{1}{4} [3 sin x - sin 3 x], {cos}^{3} x = \frac{1}{4} [cos 3 x + 3 cos x]$
we get,
$I = \frac{1}{4} \int (- sin 3 x + 3 sin x + cos 3 x + 3 cos x) d x = \frac{1}{4} [\frac{cos 3 x}{3} - 3 cos x + \frac{sin 3 x}{3} + 3 sin x] + C = \frac{3 sin x - 3 cos x}{4} + \frac{sin 3 x + cos 3 x}{12} + C$

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The value of ∫(sin3x+cos3x)dx is (where C is constant of integration)

The correct option is D 3sinx−3cosx4+sin3x+cos3x12+CLet I=∫(sin3x+cos3x)dx Using, sin3x=14[3sinx−sin3x],cos3x=14[cos3x+3cosx] we get, I=14∫(−sin3x+3sinx+cos3x+3cosx)dx=14[cos3x3−3cosx+sin3x3+3sinx]+C=3sinx−3cosx4+sin3x+cos3x12+C

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(where $C$ is constant of integration)