int frac{dx}{sin^6 x + cos^6 x} = | vedclass

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

Question

(A) We know that $\sin^6 x + \cos^6 x = (\sin^2 x + \cos^2 x)(\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x) = 1 \cdot ((\sin^2 x + \cos^2 x)^2 - 3 \sin^2 x

Vedclass · Accepted Answer

$	an^{-1} (	an x - \cot x) + C$

Vedclass · Answer

(A) We know that $\sin^6 x + \cos^6 x = (\sin^2 x + \cos^2 x)(\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x) = 1 \cdot ((\sin^2 x + \cos^2 x)^2 - 3 \sin^2 x \cos^2 x) = 1 - 3 \sin^2 x \cos^2 x$. Dividing numerator and denominator by $\cos^6 x$,we get: $\int \frac{\sec^6 x dx}{	an^6 x + 1} = \int \frac{\sec^4 x \cdot \sec^2 x dx}{	an^6 x + 1} = \int \frac{(1 + 	an^2 x)^2 \sec^2 x dx}{	an^6 x + 1}$. Let $u = 	an x$,then $du = \sec^2 x dx$. The integral becomes $\int \frac{(1 + u^2)^2}{u^6 + 1} du = \int \frac{u^4 + 2u^2 + 1}{u^6 + 1} du$. Dividing numerator and denominator by $u^3$: $\int \frac{u + \frac{2}{u} + \frac{1}{u^3}}{u^3 + \frac{1}{u^3}} du = \int \frac{u + \frac{1}{u^3} + \frac{2}{u}}{u^3 + \frac{1}{u^3}} du$. Using the substitution $t = 	an x - \cot x = u - \frac{1}{u}$,we have $dt = (1 + \frac{1}{u^2}) du = \frac{u^2 + 1}{u^2} du$. After simplification,the integral evaluates to $	an^{-1}(	an x - \cot x) + C$.

$\int \frac{dx}{\sin^6 x + \cos^6 x} = $

Similar Questions

The value of $\int \frac{2 x^{12}+5 x^9}{\left(x^5+x^3+1\right)^3} \,d x$ is equal to (where $C$ is an arbitrary constant.)

If $\int {({x^3} - 2{x^2} + 5){e^{3x}}\,dx} = e^{3x} (Ax^3 + Bx^2 + Cx + D) + K$,then the statement which is incorrect is

$\int \frac{dx}{(1+\sqrt{x}) \sqrt{x-x^2}} = $

If $\int \frac{1}{x^4+8 x^2+9} d x = \frac{1}{k} \left[ \frac{1}{\sqrt{14}} \tan^{-1}(f(x)) - \frac{1}{\sqrt{2}} \tan^{-1}(g(x)) \right] + c$, then $\sqrt{\frac{k}{2} + f(\sqrt{3}) + g(1)} =$

If $\int {\frac{{\tan x}}{{1 + \tan x + {{\tan }^2}x}}dx} = x - \frac{K}{{\sqrt A }}{\tan ^{ - 1}}\left( {\frac{{K\tan x + 1}}{{\sqrt A }}} \right) + C,$ ($C$ is a constant of integration),then the ordered pair $(K, A)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module