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

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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: $I = \int -\frac{1}{6} u^{-2/3} du$ $I = -\frac{1}{6} \cdot \frac{u^{1/3}}{1/3} + C$ $I = -\frac{1}{2} u^{1/3} + C$ $I = -\frac{1}{2} (1 + \cos^6 x)^{1/3} + C$.

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

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