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Indefinite Integrals — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals4 Marks
Question
Evaluate the following integrals:
$\int\sin^3\sqrt{\text{x}}\text{dx}$
✓

Answer

Let $\text{I}=\int\sin^3\sqrt{\text{x}}\text{dx}$
$\sqrt{\text{x}}=\text{t}$
$\text{x = t}^2$
$\text{dx}=2\text{t dt}$
$\text{I}=2\int\text{t}\sin^3\text{t dt}$
$=2\int\text{t}\Big(\frac{3\sin\text{t}-\sin3\text{t}}{4}\Big)\text{dt}$
$=\frac{1}{2}\int\text{t}(3\sin\text{t}-\sin3\text{t})\text{dt}$
Using integration by parts,
$\text{I}=\frac{1}2{}\Big[\text{t}\Big(-3\cos\text{t}+\frac{1}{3}\cos3\text{t}\Big)-\int\Big(-3\cos\text{t}+\frac{\cos3\text{t}}{3}\Big)\text{dt}\Big]$
$=\frac{1}{2}\Big[\frac{-9\text{t}\cos\text{t}+\text{t}\cos3\text{t}}{3}-\Big\{-3\sin\text{t}+\frac{\sin3\text{t}}{9}\Big\}\Big]+\text{C}$
$=\frac{1}{2}\Big[\frac{-9\text{t}\cos\text{t}+\text{t}\cos3\text{t}}{3}+\frac{27\sin\text{t}-3\sin3\text{t}}{9}\Big]+\text{C}$
$=\frac{1}{18}\big[-27\text{t}\cos\text{t}+3\text{t}\cos3\text{t}+27\sin\text{t}-3\sin3\text{t}\big]+\text{C}$
$\text{I}=\frac{1}{18}\big[3\sqrt{\text{x}}\cos3\sqrt{\text{x}}+27\sin\sqrt{\text{x}}-27\sqrt{\text{x}}\cos\sqrt{\text{x}}-3\sin3\sqrt{\text{x}}\big]+\text{C}$

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