int frac{cos x-sin x}{5+sin (2 x)} d x= | vedclass

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

Question

(B) Let $I = \int \frac{\cos x-\sin x}{5+\sin 2 x} d x$. We know that $\sin 2x = 1 - (1 - \sin 2x) = 1 - (\sin x - \cos x)^2$ is not helpful here,but

Vedclass · Accepted Answer

$\frac{1}{2} 	an ^{-1}\left[\frac{1}{2}(\sin x+\cos x)ight]+c$

Vedclass · Answer

(B) Let $I = \int \frac{\cos x-\sin x}{5+\sin 2 x} d x$. We know that $\sin 2x = 1 - (1 - \sin 2x) = 1 - (\sin x - \cos x)^2$ is not helpful here,but $(\sin x + \cos x)^2 = \sin^2 x + \cos^2 x + 2 \sin x \cos x = 1 + \sin 2x$. Thus,$\sin 2x = (\sin x + \cos x)^2 - 1$. Substituting this into the integral: $I = \int \frac{\cos x-\sin x}{5 + (\sin x + \cos x)^2 - 1} d x = \int \frac{\cos x-\sin x}{4 + (\sin x + \cos x)^2} d x$. Let $t = \sin x + \cos x$. Then $dt = (\cos x - \sin x) dx$. Substituting $t$ and $dt$ into the integral: $I = \int \frac{dt}{4 + t^2} = \frac{1}{2} 	an^{-1}\left(\frac{t}{2}ight) + C$. Substituting back $t = \sin x + \cos x$: $I = \frac{1}{2} 	an^{-1}\left(\frac{\sin x + \cos x}{2}ight) + C$.

$\int \frac{\cos x-\sin x}{5+\sin (2 x)} d x=$

Similar Questions

$\int \frac{\log \sqrt{x}}{3 x} d x$ is equal to

If $f(x) = \int \frac{\sin 2x + 2 \cos x}{4 \sin^2 x + 5 \sin x + 1} \, dx$ and $f(0) = 0$, then $f\left(\frac{\pi}{6}\right) =$

If $\int \frac{\sin x}{\sin (x-\alpha)} dx = Ax + B \log |\sin (x-\alpha)| + c$,then the values of $A$ and $B$ are respectively (where $c$ is a constant of integration).

Let $f(x) = \frac{x}{(1+x^n)^{1/n}}$ for $n \geq 2$ and $g(x) = \underbrace{(f \circ f \circ \ldots \circ f)}_{n \text{ times }}(x)$. Then $\int x^{n-2} g(x) \, dx$ equals

$\int (x+1)(x+3)(x+2)^7 \, dx = $ . . . . . . $+ C$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module