Question

If $\underset{0}{\overset{4}{\int }}\frac{dx}{x+\sqrt{16-x^2}}=k,$ then which of the following is/are CORRECT?

• A. $4k=1$
• B. $4k=\pi$
• C. $2k=1$
• D. $\cos \left(2k\right)=0$

Answer 1

Answer: (B), (D)

$\cos \left(2k\right)=0$

Given : $k=\underset{0}{\overset{4}{\int }}\frac{dx}{x+\sqrt{16-x^2}}$
Put, $x=4\sin y$
$\Rightarrow dx=4\left(\cos y\right)dy$
when, $x=0\Rightarrow y=0;$
when, $x=4\Rightarrow y=\frac{\pi }{2}$
Now, integral becomes :
$k=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{4\cos y}{4\sin y+4\cos y}dy=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{\cos y}{\sin y+\cos y}dy\cdots \left(i\right)$
Using the property:
$\left[\underset{a}{\overset{b}{\int }}f\left(x\right)dx=\underset{a}{\overset{b}{\int }}f(a+b-x)dx\right]$
$k=\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{\sin y}{\sin y+\cos y}dy\cdots \left(ii\right)$
Adding $\left(i\right)$ and $\left(ii\right):$
$\Rightarrow 2k=\underset{0}{\overset{\frac{\pi }{2}}{\int }}dy=\frac{\pi }{2}$
So, $k=\frac{\pi }{4}$