If int frac{sin x cos x}{sqrt{cos^4 x - sin^4 | vedclass

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

Question

(C) Given integral $I = \int \frac{\sin x \cos x}{\sqrt{\cos^4 x - \sin^4 x}} dx$.Using $\sin 2x = 2 \sin x \cos x$ and $\cos^4 x - \sin^4 x = (\cos^2

Vedclass · Accepted Answer

$[(4n-1)\frac{\pi}{4}, (4n+1)\frac{\pi}{4}], n=0, 1, 2, \ldots$

Vedclass · Answer

(C) Given integral $I = \int \frac{\sin x \cos x}{\sqrt{\cos^4 x - \sin^4 x}} dx$.Using $\sin 2x = 2 \sin x \cos x$ and $\cos^4 x - \sin^4 x = (\cos^2 x - \sin^2 x)(\cos^2 x + \sin^2 x) = \cos 2x$.So,$I = \int \frac{\frac{1}{2} \sin 2x}{\sqrt{\cos 2x}} dx = \frac{1}{2} \int \sin 2x (\cos 2x)^{-1/2} dx$.Let $t = \cos 2x$,then $dt = -2 \sin 2x dx$,so $\sin 2x dx = -\frac{1}{2} dt$.$I = \frac{1}{2} \int -\frac{1}{2} t^{-1/2} dt = -\frac{1}{4} \cdot 2 t^{1/2} + c = -\frac{1}{2} \sqrt{\cos 2x} + c$.Comparing with $-\frac{f(x)}{2} + c$,we get $f(x) = \sqrt{\cos 2x}$.For the domain,$\cos 2x \geq 0$.This implies $2n\pi - \frac{\pi}{2} \leq 2x \leq 2n\pi + \frac{\pi}{2}$.Dividing by $2$,we get $n\pi - \frac{\pi}{4} \leq x \leq n\pi + \frac{\pi}{4}$.For $n=0$,$[-\frac{\pi}{4}, \frac{\pi}{4}]$. For $n=1$,$[\frac{3\pi}{4}, \frac{5\pi}{4}]$.This matches the general form $[(4n-1)\frac{\pi}{4}, (4n+1)\frac{\pi}{4}]$.

If $\int \frac{\sin x \cos x}{\sqrt{\cos^4 x - \sin^4 x}} dx = -\frac{f(x)}{2} + c$,then the domain of $f(x)$ is

Similar Questions

$\int \sqrt{x+\sqrt{x^2+2}} \, dx =$

Integrate the function: $\frac{1}{\sqrt{(x-a)(x-b)}}$

$\int \tan ^{-1}\left(1-x+x^2\right) d x+\int \tan ^{-1}(x) d x+\int \tan ^{-1}(1-x) d x=$

For any integer $n \geq 2$, let $I_n = \int \tan^n x \, dx$. If $I_n = \frac{1}{a} \tan^{n-1} x - b I_{n-2}$ for $n \geq 2$, then the ordered pair $(a, b)$ is equal to

Given that $\int \frac{1}{x^2+a^2} dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C$. If $\int \frac{1}{x^4+3x^2+1} dx = a \cdot \tan^{-1}\left(\frac{b(x^2-1)}{x}\right) + c \cdot \tan^{-1}\left(\frac{d(x^2+1)}{x}\right) + k$, where $k$ is a constant of integration, then $5(c+d+ab) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module