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Indefinite Integrals — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsIndefinite Integrals5 Marks
Question
Evaluate the following integrals:
$\int\frac{\sin^3\text{x}}{\sqrt{\cos\text{x}}}\text{dx}$
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Answer

Let I $=\int\frac{\sin^3\text{x}}{\sqrt{\cos\text{x}}}\text{dx}$$\therefore\text{I}=\int\frac{\sin^2\text{x}\sin\text{x}}{\sqrt{\cos\text{x}}}\text{dx}$
$\Rightarrow\text{I}=\int\frac{(1-\cos^2\text{x})}{\sqrt{\cos\text{x}}}\sin\text{x dx}\ ...(1)$
Let $\cos\text{x}=\text{t}$ then, $\text{d}(\cos\text{x})=\text{dt}$
$\Rightarrow-\sin\text{x dx}=\text{dt}$
$\Rightarrow\sin\text{x dx}=-\text{dt}$
Putting $\cos\text{x}=\text{t}$ and $\sin\text{x dx}=-\text{dt}$ in equation (1), we get
$\text{I}=\int\frac{(1-\text{t}^2)}{\sqrt{\text{t}}}\times-\text{dt}$
$=\int\frac{\text{t}^2-1}{\sqrt{\text{t}}}\text{dt}$
$=\int\Bigg(\frac{\text{t}^2}{\text{t}^\frac{1}{2}}-\frac{1}{\text{t}^\frac{1}{2}}\Bigg)\text{dt}$
$=\int\Big(\text{t}^{2-\frac{1}{2}}-\text{t}^\frac{-1}{2}\Big)\text{dt}$
$=\int\Big(\text{t}^\frac{3}{2}-\text{t}^{\frac{-1}{2}}\Big)\text{dt}$
$=\frac{2}{5}\text{t}^\frac{5}{2}-2\text{t}^\frac{1}{2}+\text{C}$
$\therefore\text{I}=\frac{2}{5}\cos^\frac{5}{2}\text{x}-2\cos^\frac{1}{2}\text{x}+\text{C}$
$\therefore\text{I}=\frac{2}{5}\cos^\frac{5}{2}\text{x}-2\sqrt{\cos\text{x}}+\text{C}$

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