# ∫ Sin 3 √ X D X - Mathematics

Sum
$\int \sin^3 \sqrt{x}\ dx$

#### Solution

$\text{ Let, }I = \int \sin^3 \sqrt{x} dx . . . . . \left( 1 \right)$
$\text{ Consider,} \sqrt{x} = t . . . . . \left( 2 \right)$
$\text{Differentiating both sides we get},$
$\frac{1}{2\sqrt{x}}dx = dt$
$\Rightarrow dx = 2\sqrt{x} \text{ dt }$
$\Rightarrow dx = \text{ 2t dt }$
$\text{ Therefore,} \left( 1 \right) \text{ becomes,}$
$I = \int \sin^3 t \text{ 2t dt }$
$= 2\int t \sin^3 \text{ t dt }$
$= 2\int t \left( \frac{3\sin t - \sin 3t}{4} \right) dt \left( \text{ Since, }\sin 3A = 3\sin A - 4 \sin^3 A \right)$
$= \frac{3}{2}\int \text{ t sin t dt } - \frac{1}{2}\int t \text{ sin 3t dt }$
$= \frac{3}{2}\left[ t\int\text{ sin t dt }- \int\left( \frac{d t}{d t}\int\text{ sin t dt } \right)dt \right] - \frac{1}{2}\left[ t\int \text{ sin 3t dt }- \int\left( \frac{d t}{d t}\int\text{ sin 3t dt } \right)dt \right]$
$= \frac{3}{2}\left[ - \text{ t cos t } + \int\text{ cos t dt }\right] - \frac{1}{2}\left[ - \frac{t \cos 3t}{3} + \frac{1}{3}\int\text{ cos 3t dt }\right]$
$= \frac{3}{2}\left[ - t \cos t + \sin t \right] - \frac{1}{2}\left[ - \frac{t \cos3t}{3} + \frac{1}{9}\text{ sin 3t }\right] + C$
$= - \frac{3}{2}t \cos t + \frac{3}{2}\sin t + \frac{1}{6}t \text{ cos 3t} - \frac{1}{18}\text{ sin 3t} + C$
$= - \frac{3}{2}\sqrt{x}\cos\sqrt{x} + \frac{3}{2}\sin\sqrt{x} + \frac{1}{6}\sqrt{x}\cos\left( 3\sqrt{x} \right) - \frac{1}{18}\sin\left( 3\sqrt{x} \right) + C$

Concept: Indefinite Integral Problems
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#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 19 Indefinite Integrals
Exercise 19.25 | Q 53 | Page 134

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