Question

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

Answer 1

Here $I=\int {\cos }^3\sqrt{x}dx$

Let $\sqrt{x}=\theta$

Differetiate on both side

$\frac{1}{2\sqrt{x}}dx=d\theta$

$dx=2\sqrt{x}d\theta$

$dx=2\theta d\theta$

put the value of dx in integeral

$I=\int {\cos }^3\theta \left(2\theta d\theta \right)$

As we know that the expansion

$\cos 3\theta =4{\cos }^3\theta -3\cos \theta$

${\cos }^3\theta =\frac{\cos 3\theta +3\cos \theta }{4}$

Finally $I$ can be written as:-

$I=\frac{1}{2}\int \theta (\cos 3\theta +3\cos \theta )d\theta$

$=\frac{1}{2}[\theta \int (\cos 3\theta +\cos \theta )d\theta -3(\int \frac{d\left(\theta \right)}{d\theta }\int (\cos 3\theta +\cos \theta d\theta \left)d\theta \right]$

$=\frac{1}{2}\left[\theta \right(\frac{\sin 3\theta }{3}+\sin \theta )-3\{\int (\frac{\sin 3\theta }{3}+\sin \theta )d\theta \left\}\right]$

$=\frac{1}{2}\left[\theta \right(\frac{\sin 3\theta }{3}+\sin \theta )-3\{\frac{(-)\cos 3\theta }{9}+(-)\cos \theta \left\}\right]+C$

$I=\frac{1}{6}[\theta \sin 3\theta +3\theta \sin \theta +\cos 3\theta +9\cos \theta ]+C$

Where C is integeral constant