Question

If $\int \frac{cosx-sinx}{\sqrt{8-sin2x}}dx=si{n}^{-1}\left(\frac{A}{426}\right(sinx+cosx\left)\right)+C$ then A is equal to

Answer 1

Let $I=\int \frac{\cos x-\sin x}{\sqrt{8-\sin 2x}}dx$

$I=\int \frac{\cos x-\sin x}{\sqrt{9-1-\sin 2x}}dx$

$I=\int \frac{\cos x-\sin x}{\sqrt{9-({\sin }^{2}2x+{\cos }^{2}2x)-\sin 2x}}dx$

$=\int \frac{\cos x-\sin x}{\sqrt{9-{\left(\sin x+\cos x\right)}^2}}dx$

Substituting $\sin x+\cos x=t\Rightarrow (\cos x-\sin x)dx=dt$

$\sin x+\cos x=t\Rightarrow (\cos x-\sin x)dx=dt$

$I=\int \frac{dt}{\sqrt{9-t^2}}={\sin }^{-1}\frac{t}{3}+c$

We know that, $\int \frac{1}{\sqrt{a^2-x^2}}dx={\sin }^{-1}\left(\frac{x}{a}\right)+c$

$I={\sin }^{-1}\frac{\left(\sin x+\cos x\right)}{3}+c$

Comparing with $\int \frac{cosx-sinx}{\sqrt{8-sin2x}}dx=si{n}^{-1}\left(\frac{A}{426}\right(sinx+cosx\left)\right)+C$

Therefore, $A=142$