Question

If ${\int }_{}\frac{\cos x-\sin x}{\sqrt{8-\sin 2x}}dx=a{\sin }^{-1}\left(\frac{\sin x+\cos x}{b}\right)+c$ where $c$ is a constant of integration, then the ordered pair $\left(a,b\right)$ is equal to:

• A. $\left(1,-3\right)$
• B. $\left(1,3\right)$
• C. $\left(-1,3\right)$
• D. $\left(3,1\right)$

Answer 1

The correct option is B

$\left(1,3\right)$

Explanation for the correct option.

Step 1. Form the required substitution.

Let $\sin x+\cos x=t$, then on differentiating it is found that $\left(\cos x-\sin x\right)dx=dt$.

Now, square both sides of the equation $\sin x+\cos x=t$.

$$\begin{gathered} {\left(\sin x+\cos x\right)}^2=t^2 \\ \Rightarrow {\sin }^{2}x+{\cos }^{2}x+2\sin x\cos x=t^2 \\ \Rightarrow 1+\sin 2x=t^2 \\ \Rightarrow \sin 2x=t^2-1 \end{gathered}$$

Step 2. Find the ordered pair $\left(a,b\right)$.

In the integral ${\int }_{}\frac{\cos x-\sin x}{\sqrt{8-\sin 2x}}dx$, substitute $\left(\cos x-\sin x\right)dx=dt$ and $\sin 2x=t^2-1$.

$\begin{array}{rcl}{\int }_{}\frac{\cos x-\sin x}{\sqrt{8-\sin 2x}}dx& =& {\int }_{}\frac{dt}{\sqrt{8-\left(t^2-1\right)}}\\ & =& {\int }_{}\frac{dt}{\sqrt{9-t^2}}\\ & =& {\sin }^{-1}\left(\frac{t}{3}\right)+c\\ & =& {\sin }^{-1}\left(\frac{\sin x+\cos x}{3}\right)+c\left[t=sinx+cosx\right]\end{array}$

On comparing $\begin{array}{rcl}& & {\sin }^{-1}\left(\frac{\sin x+\cos x}{3}\right)+c\end{array}$ with $a\sin x$ it is found that $a=1,b=3$.

So the ordered pair $\left(a,b\right)$ is equal to $\left(1,3\right)$.

Hence, the correct option is B.