int frac{sin ^8 x-cos ^8 x}{1-2 sin ^2 x cos | vedclass

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

Question

(D) We have the integral $I = \int \frac{\sin ^8 x-\cos ^8 x}{1-2 \sin ^2 x \cos ^2 x} d x$. Using the identity $a^2 - b^2 = (a-b)(a+b)$, we can facto

Vedclass · Accepted Answer

$\frac{-1}{2} \sin 2 x+c$

Vedclass · Answer

(D) We have the integral $I = \int \frac{\sin ^8 x-\cos ^8 x}{1-2 \sin ^2 x \cos ^2 x} d x$. Using the identity $a^2 - b^2 = (a-b)(a+b)$, we can factor the numerator: $\sin ^8 x - \cos ^8 x = (\sin ^4 x - \cos ^4 x)(\sin ^4 x + \cos ^4 x) = (\sin ^2 x - \cos ^2 x)(\sin ^2 x + \cos ^2 x)(\sin ^4 x + \cos ^4 x)$. Since $\sin ^2 x + \cos ^2 x = 1$, the numerator becomes $(\sin ^2 x - \cos ^2 x)(\sin ^4 x + \cos ^4 x)$. Now, consider the denominator: $1 - 2 \sin ^2 x \cos ^2 x$. We know that $1 = (\sin ^2 x + \cos ^2 x)^2 = \sin ^4 x + \cos ^4 x + 2 \sin ^2 x \cos ^2 x$. Therefore, $1 - 2 \sin ^2 x \cos ^2 x = \sin ^4 x + \cos ^4 x$. Substituting these into the integral: $I = \int \frac{(\sin ^2 x - \cos ^2 x)(\sin ^4 x + \cos ^4 x)}{\sin ^4 x + \cos ^4 x} d x = \int (\sin ^2 x - \cos ^2 x) d x$. Using the identity $\cos 2x = \cos ^2 x - \sin ^2 x$, we get $\sin ^2 x - \cos ^2 x = -\cos 2x$. Thus, $I = \int -\cos 2x d x = -\frac{1}{2} \sin 2x + C$.

$\int \frac{\sin ^8 x-\cos ^8 x}{1-2 \sin ^2 x \cos ^2 x} d x=$

Similar Questions

$\int \frac{dx}{a^2 \sin^2 x + b^2 \cos^2 x}$ is equal to

If $\int(x^6+x^4+x^2) \sqrt{2x^4+3x^2+6} dx = f(x) + c$,then $f(3) =$

$ \int \frac{1}{x^{2}\left(x^{4}+1\right)^{3 / 4}} dx $ is equal to

If the value of the integral $\int_{1}^{2} e^{x^2} dx$ is $\alpha$,then the value of $\int_{e}^{e^4} \sqrt{\ln x} dx$ is:

If $\int \frac{1}{\cos^3 x \sqrt{\sin 2x}} dx = p(\tan^2 x + q)\sqrt{\tan x} + c$, then the values of $p$ and $q$ respectively are

Vedclass Test Series

Exam Paper Generator

Online Exam Module