If int sec ^4 x cdot tan ^4 x , dx = frac{tan | vedclass

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(B) Let $I = \int \sec ^4 x 	an ^4 x \, dx$. We can rewrite the integral as $I = \int \sec ^2 x \cdot \sec ^2 x \cdot 	an ^4 x \, dx$. Using the ide

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$12$

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(B) Let $I = \int \sec ^4 x 	an ^4 x \, dx$. We can rewrite the integral as $I = \int \sec ^2 x \cdot \sec ^2 x \cdot 	an ^4 x \, dx$. Using the identity $\sec ^2 x = 1 + 	an ^2 x$,we get $I = \int (1 + 	an ^2 x) 	an ^4 x \cdot \sec ^2 x \, dx$. Let $	an x = t$,then $\sec ^2 x \, dx = dt$. Substituting these into the integral,we get $I = \int (1 + t^2) t^4 \, dt = \int (t^4 + t^6) \, dt$. Integrating with respect to $t$,we obtain $I = \frac{t^5}{5} + \frac{t^7}{7} + c$. Substituting back $t = 	an x$,we have $I = \frac{	an ^5 x}{5} + \frac{	an ^7 x}{7} + c$. Comparing this with the given form $\frac{	an ^m x}{m} + \frac{	an ^n x}{n} + c$,we get $m = 5$ and $n = 7$ (or vice versa). Therefore,$m + n = 5 + 7 = 12$.

If $\int \sec ^4 x \cdot \tan ^4 x \, dx = \frac{\tan ^m x}{m} + \frac{\tan ^n x}{n} + c$ (where $c$ is the constant of integration),then $m + n =$

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