If int frac{cos x-sin x}{sqrt{8-sin 2 x}} ,d | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $I = \int \frac{\cos x - \sin x}{\sqrt{8 - \sin 2x}} dx$. We know that $\sin 2x = 1 - (1 - \sin 2x) = 1 - (\sin x - \cos x)^2$,but it is more

Vedclass · Accepted Answer

$(1, 3)$

Vedclass · Answer

(A) Let $I = \int \frac{\cos x - \sin x}{\sqrt{8 - \sin 2x}} dx$. We know that $\sin 2x = 1 - (1 - \sin 2x) = 1 - (\sin x - \cos x)^2$,but it is more convenient to write $8 - \sin 2x = 9 - (1 + \sin 2x) = 9 - (\sin x + \cos x)^2$. So,$I = \int \frac{\cos x - \sin x}{\sqrt{9 - (\sin x + \cos x)^2}} dx$. Let $t = \sin x + \cos x$. Then $dt = (\cos x - \sin x) dx$. Substituting these into the integral,we get $I = \int \frac{dt}{\sqrt{3^2 - t^2}}$. Using the standard integral formula $\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1}(\frac{x}{a}) + c$,we have $I = \sin^{-1}(\frac{t}{3}) + c$. Substituting $t = \sin x + \cos x$ back,we get $I = \sin^{-1}\left(\frac{\sin x + \cos x}{3}\right) + c$. Comparing this with the given expression $a \sin^{-1}\left(\frac{\sin x + \cos x}{b}\right) + c$,we find $a = 1$ and $b = 3$. Thus,the ordered pair $(a, b)$ is $(1, 3)$.

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

Similar Questions

$\int {\frac{{{e^{\sqrt x }}}}{{\sqrt x }}dx} = $

The integral $\int \frac{\sin ^2 x \cos ^2 x}{\left(\sin ^5 x+\cos ^3 x \sin ^2 x+\sin ^3 x \cos ^2 x+\cos ^5 x\right)^2} d x$ is equal to (Where $C$ is a constant of integration).

$\int \frac{d x}{\sin x+\cos x}=$

$\int \frac{\cos \sqrt{x}}{\sqrt{x}} dx = $

$\int \frac{\sin ^6 x}{\cos ^8 x} d x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module