Let $n$ be the number of integral values of $x$, which satisfy the equation $ an ^{-1} x-\sin ^{-1}

Question

Let $n$ be the number of integral values of $x$, which satisfy the equation $	an ^{-1} x-\sin ^{-1} x=\cos ^{-1} x-\cot ^{-1} x$.
Let $\vec{c}=|\vec{a} 	imes \vec{b}| \vec{a}+(\vec{a} \cdot \vec{b}) \vec{b} \quad$ where $\vec{a}$ and $\vec{b}$ are unit vectors and angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{6}$
Let the value of definite integral $\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{\ln (\sec x)}{(2013)^{\sin x}+1} d x$ equals $\frac{\pi}{2} \ln k$, then the value of $\left(\mathrm{n}+2|\overrightarrow{\mathrm{c}}|^2+\mathrm{k}ight) \quad$ is equal to....

Professor PC Thomas Classes · Answer

We have $	an ^{-1} x+\cot ^{-1} x=\sin ^{-1} x+\cos ^{-1} x$, which is true for all $x \in[-1,1]$ So, number of integral values of $\mathbf{x}=\mathbf{3}$.
$$
\begin{aligned}
& \vec{c}=|\vec{a} 	imes \vec{b}| \vec{a}+(\vec{a} \cdot \vec{b}) \vec{b},|\vec{a}|=1=|\vec{b}| \
& \vec{a} \wedge \vec{b}=\frac{\pi}{6} \
& 	herefore|\vec{a} 	imes \vec{b}|=(1)(1) \sin \frac{\pi}{6}=\frac{1}{2} \vec{a} \cdot \vec{b}=(1)(1) \
& \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2} \
& \Rightarrow \vec{c}=\frac{1}{2} \vec{a}+\frac{\sqrt{3}}{2} \vec{b} \
& \Rightarrow|2 \vec{c}|^2=|\vec{a}+\sqrt{3} \vec{b}|^2 \
& \Rightarrow 4|\vec{c}|^2=1+3+2 \sqrt{3}\left(\frac{\sqrt{3}}{2}ight)=7 \
& 	herefore 4|\vec{c}|^2=7 \
& 	ext { Let } I=\int_{-\pi / 2}^{\pi / 2} \frac{\ln (\sec x)}{(2013)^{\sin x}+1} d x
\end{aligned}
$$
So,
$$
\begin{aligned}
& {\left[\frac{1}{(2013)^{\sin x}+1}+\frac{1}{(2013)^{-\sin x}+1}ight] d x} \
& \quad=-\int_0^{\pi / 2} \ell \ln (\cos x) d x=\frac{\pi}{2} \ell n 2 \
& 	herefore k=2
\end{aligned}
$$