int sec^{frac{2}{3}} x cdot operatorname{cose | vedclass

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

Question

(A) Let $I = \int \sec^{\frac{2}{3}} x \cdot \operatorname{cosec}^{\frac{4}{3}} x \, dx$. We can rewrite the integrand as: $I = \int \frac{1}{\cos^{\f

Vedclass · Accepted Answer

$-3 	an^{-\frac{1}{3}} x + c$,where $c$ is the constant of integration

Vedclass · Answer

(A) Let $I = \int \sec^{\frac{2}{3}} x \cdot \operatorname{cosec}^{\frac{4}{3}} x \, dx$. We can rewrite the integrand as: $I = \int \frac{1}{\cos^{\frac{2}{3}} x} \cdot \frac{1}{\sin^{\frac{4}{3}} x} \, dx$. Divide the numerator and denominator by $\cos^{\frac{4}{3}} x$: $I = \int \frac{1}{\cos^{\frac{2}{3}} x \cdot \sin^{\frac{4}{3}} x} \cdot \frac{\cos^{\frac{4}{3}} x}{\cos^{\frac{4}{3}} x} \, dx = \int \frac{\cos^{\frac{4}{3}} x}{\cos^2 x \cdot \sin^{\frac{4}{3}} x} \, dx$. $I = \int \frac{1}{\cos^2 x} \cdot \left( \frac{\sin x}{\cos x} ight)^{-\frac{4}{3}} \, dx = \int \sec^2 x \cdot (	an x)^{-\frac{4}{3}} \, dx$. Let $u = 	an x$,then $du = \sec^2 x \, dx$. $I = \int u^{-\frac{4}{3}} \, du = \frac{u^{-\frac{4}{3} + 1}}{-\frac{4}{3} + 1} + c = \frac{u^{-\frac{1}{3}}}{-\frac{1}{3}} + c = -3u^{-\frac{1}{3}} + c$. Substituting $u = 	an x$ back,we get: $I = -3 	an^{-\frac{1}{3}} x + c$.

$\int \sec^{\frac{2}{3}} x \cdot \operatorname{cosec}^{\frac{4}{3}} x \, dx =$

Similar Questions

The value of $\int \frac{x^{2}-1}{x^{4}+3 x^{2}+1} d x$ for $x>0$ is

$\int \sqrt{\frac{\cos x - \cos^3 x}{1 - \cos^3 x}} \, dx$ is equal to

If $\int \sqrt{x}(1-x^3)^{-\frac{1}{2}} dx = \frac{2}{3} g(f(x)) + c$,then

If $\int \cos ^{\frac{3}{5}} x \cdot \sin ^3 x \,d x = \frac{-1}{m} \cos ^{m} x + \frac{1}{n} \cos ^{n} x + c$,(where $c$ is the constant of integration),then $(m, n) = $

If $\int \frac{x^2}{\sqrt{1-x}} \,d x = p \sqrt{1-x} (3x^2 + 4x + 8) + c$ where $c$ is a constant of integration, then the value of $p$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module