Find the integral. $$\int \tan ^ { 2 } x \sec ^ { 6 } x d x$$
A) $\frac { 1 } { 7 } \tan ^ { 7 } x + \frac { 2 } { 5 } \tan ^ { 5 } x + \frac { 1 } { 3 } \tan ^ { 3 } x + C$
B) $\frac { 1 } { 5 } \tan ^ { 5 } x + \frac { 2 } { 3 } \tan ^ { 3 } x + \tan x + C$
C) $\frac { 1 } { 5 } \tan ^ { 5 } x + \frac { 2 } { 3 } \tan ^ { 3 } x - \tan x + C$
D) $\frac { 1 } { 7 } \tan ^ { 7 } x + \frac { 2 } { 5 } \tan ^ { 5 } x - \frac { 1 } { 3 } \tan ^ { 3 } x + C$

Question

Quizplus · Accepted Answer

The integral of $\tan^2x \sec^6x dx$ can be solved by recognizing that $\tan^2x = \sec^2x - 1$ and using substitution or direct integration techniques, leading to the answer which involves terms of $\tan^7x$, $\tan^5x$, and $\tan^3x$, with the coefficients matching those in option A.

Find the Integral ∫tan⁡2xsec⁡6xdx\int \tan ^ { 2 } x \sec ^ { 6 } x d x∫tan2xsec6xdx

Find the Integral $\int \tan ^ { 2 } x \sec ^ { 6 } x d x$