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

$\int \frac{x^4+1}{x^6+1} dx = $

Evaluate the integral $\int \frac{x^2 + \cos^2 x}{1 + x^2} \operatorname{cosec}^2 x \, dx$,where $c$ is the constant of integration.

$\int \sqrt{3-2x-x^2} \, dx = $ . . . . . . $+ C$.

If $n$ is a positive integer greater than $1$ and $I_{n}=\int \frac{\sin n x}{\sin x} d x$, then $I_{n+1}-I_{n-1}=$

If $\int \frac{4e^x + 6e^{-x}}{9e^x - 4e^{-x}} dx = Ax + B \log |9e^{2x} - 4| + C$,then $A, B$ and $C$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module