Question

$\int {\cos }^3\sqrt{x}dx$

Answer 1

$$\begin{gathered} \mathrm{Let},I=\int {\cos }^3\sqrt{x}dx.....\left(1\right) \\ \mathrm{Consider},\sqrt{x}=t.....\left(2\right) \\ \mathrm{Differentiating}\mathrm{both}\mathrm{sides}\mathrm{we}\mathrm{get}, \\ \frac{1}{2\sqrt{x}}dx=dt \\ \Rightarrow dx=2\sqrt{x}dt \\ \Rightarrow dx=2tdt \\ \mathrm{Therefore},\left(1\right)\mathrm{becomes}, \\ I=\int {\cos }^{3}t2tdt \\ =2\int t{\cos }^{3}tdt \\ =2\int t\left(\frac{3\cos t+\cos 3t}{4}\right)dt\left(\mathrm{Since},\cos 3A=4{\cos }^{3}A-3\cos A\right) \\ =\frac{3}{2}\int t\cos tdt+\frac{1}{2}\int t\cos 3tdt \\ =\frac{3}{2}\left[t\int \cos tdt-\int \left(\frac{\mathit{d}t}{\mathit{d}t}\int \cos tdt\right)dt\right]+\frac{1}{2}\left[t\int \cos 3tdt-\int \left(\frac{\mathit{d}t}{\mathit{d}t}\int \cos 3tdt\right)dt\right] \\ =\frac{3}{2}\left[t\sin t-\int \sin tdt\right]+\frac{1}{2}\left[\frac{t\sin 3t}{3}-\frac{1}{3}\int \sin 3tdt\right] \\ =\frac{3}{2}\left[t\sin t+\cos t\right]+\frac{1}{2}\left[\frac{t\sin 3t}{3}+\frac{1}{9}\cos 3t\right]+C \\ =\frac{3}{2}t\sin t+\frac{3}{2}\cos t+\frac{1}{6}t\sin 3t+\frac{1}{18}\cos 3t+C \\ =\frac{3}{2}\sqrt{x}\sin \sqrt{x}+\frac{3}{2}\cos \sqrt{x}+\frac{1}{6}\sqrt{x}\sin \left(3\sqrt{x}\right)+\frac{1}{18}\cos \left(3\sqrt{x}\right)+C \end{gathered}$$

Note: The final answer in indefinite integration may vary based on the integration constant.