Question

If $\underset{0}{\overset{1}{\int }}{x}^6\sqrt{1-x^2}dx=\frac{\lambda \pi }{k}$ then find $\lambda +k$

Answer 1

Answer: 261.00

Let $x=\sin \theta \Rightarrow dx=\cos \theta d\theta$
$\underset{0}{\overset{\pi /2}{\int }}{\sin }^6\theta \cdot {\cos }^2\theta d\theta$
$=\frac{5\cdot 3\cdot 1\cdot 1}{8\cdot 6\cdot 4\cdot 2}\times \frac{\pi }{2}=\frac{5\pi }{256}$