int frac{tan x}{sec ^2 xleft(1+sec ^6 xright) | vedclass

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

Question

(C) Let $I = \int \frac{	an x}{\sec ^2 x(1+\sec ^6 x)^{2/3}} dx$. Rewrite the integrand as: $I = \int \frac{	an x}{\sec ^2 x \cdot (\sec^6 x)^{2/3}

Vedclass · Accepted Answer

$\frac{-1}{2}\left(1+\cos ^6 x\right)^{\frac{1}{3}}+c$

Vedclass · Answer

(C) Let $I = \int \frac{	an x}{\sec ^2 x(1+\sec ^6 x)^{2/3}} dx$. Rewrite the integrand as: $I = \int \frac{	an x}{\sec ^2 x \cdot (\sec^6 x)^{2/3} (\frac{1}{\sec^6 x} + 1)^{2/3}} dx$ $I = \int \frac{	an x}{\sec ^2 x \cdot \sec^4 x (\cos^6 x + 1)^{2/3}} dx$ $I = \int \frac{	an x}{\sec^6 x (1 + \cos^6 x)^{2/3}} dx$ $I = \int \frac{\sin x}{\cos x} \cdot \cos^6 x (1 + \cos^6 x)^{-2/3} dx$ $I = \int \sin x \cos^5 x (1 + \cos^6 x)^{-2/3} dx$. Let $u = 1 + \cos^6 x$. Then $du = 6 \cos^5 x (-\sin x) dx$, which implies $\sin x \cos^5 x dx = -\frac{1}{6} du$. Substituting these into the integral: $I = \int -\frac{1}{6} u^{-2/3} du$ $I = -\frac{1}{6} \cdot \frac{u^{1/3}}{1/3} + C$ $I = -\frac{1}{2} u^{1/3} + C$ $I = -\frac{1}{2} (1 + \cos^6 x)^{1/3} + C$.

$\int \frac{\tan x}{\sec ^2 x\left(1+\sec ^6 x\right)^{\frac{2}{3}}} d x=$

Similar Questions

$\int \frac{x^4 \cos \left(\tan ^{-1} x^5\right)}{1+x^{10}} \,d x$ equals

If $\int \frac{\sqrt{x}}{x(x+1)} dx = k \tan^{-1} m + c$,(where $c$ is the constant of integration),then:

$\int \frac{2 \tan (x)}{1+2 \tan ^2(x)} d x=$

$\int \frac{1}{x} \sec^2(\log x) \, dx = $

$\int \frac{1}{3-2 \cos 2 x} \,d x=$ (where $C$ is constant of integration.)

Vedclass Test Series

Exam Paper Generator

Online Exam Module