Question

If $I_n=\underset{0}{\overset{1}{\int }}({\cos }^{-1}x{)}^{n}dx,$ then $I_4$ is equal to

• A. $\frac{{\pi }^3}{2}-12\pi -24$
• B. $\frac{{\pi }^3}{2}-12\pi +24$
• C. $\frac{{\pi }^3}{2}-12\pi$
• D. $\frac{{\pi }^3}{4}-12\pi +24$

Answer 1

The correct option is B $\frac{{\pi }^3}{2}-12\pi +24$

$I_n=\underset{0}{\overset{1}{\int }}({\cos }^{-1}x{)}^{n}dx$
Put $x=\cos \theta \Rightarrow dx=-\sin \theta d\theta$
$\Rightarrow I_n=\underset{0}{\overset{\pi /2}{\int }}{\theta }^n\cdot \sin \theta d\theta$
Applying by parts
$=[{\theta }^n\cdot (-\cos \theta ){]}_0^{\pi /2}+n\underset{0}{\overset{\pi /2}{\int }}{\theta }^{n-1}\cos \theta d\theta$
$=0+n{[{\theta }^{n-1}\cdot \sin \theta -(n-1\left)\int {\theta }^{n-2}\cdot \sin \theta d\theta \right]}_0^{\pi /2}$
$=[n\cdot {\theta }^{n-1}\cdot \sin \theta {]}_0^{\pi /2}-n(n-1)\underset{0}{\overset{\pi /2}{\int }}{\theta }^{n-2}\cdot \sin \theta d\theta$
$\Rightarrow I_n=n{\left(\frac{\pi }{2}\right)}^{n-1}-n(n-1)I_{n-2}$

Now, $I_4=4{\left(\frac{\pi }{2}\right)}^3-12I_2$
$\Rightarrow I_4=4{\left(\frac{\pi }{2}\right)}^3-12[2\left(\frac{\pi }{2}\right)-2I_0]$
$\Rightarrow I_4=\frac{{\pi }^3}{2}-12\pi +24$