int (tan^7 x + tan x) dx = | vedclass

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(A) Let $I = \int (	an^7 x + 	an x) dx$. We can factor out $	an x$: $I = \int 	an x (	an^6 x + 1) dx$. Using the identity $	an^2 x = \sec^2 x -

Vedclass · Accepted Answer

$\frac{	an^2 x}{12} (2 	an^4 x - 3 	an^2 x + 6) + c$

Vedclass · Answer

(A) Let $I = \int (	an^7 x + 	an x) dx$. We can factor out $	an x$: $I = \int 	an x (	an^6 x + 1) dx$. Using the identity $	an^2 x = \sec^2 x - 1$,we have $	an^6 x = (\sec^2 x - 1)^3$. Alternatively,simplify the expression as follows: $I = \int (	an^7 x + 	an^5 x - 	an^5 x - 	an^3 x + 	an^3 x + 	an x) dx$ $I = \int (	an^5 x(	an^2 x + 1) - 	an^3 x(	an^2 x + 1) + 	an x(	an^2 x + 1)) dx$ Since $1 + 	an^2 x = \sec^2 x$: $I = \int (	an^5 x \sec^2 x - 	an^3 x \sec^2 x + 	an x \sec^2 x) dx$. Let $u = 	an x$,then $du = \sec^2 x dx$. $I = \int (u^5 - u^3 + u) du = \frac{u^6}{6} - \frac{u^4}{4} + \frac{u^2}{2} + C$. Factor out $\frac{u^2}{12}$: $I = \frac{u^2}{12} (2u^4 - 3u^2 + 6) + C$. Substituting $u = 	an x$ back: $I = \frac{	an^2 x}{12} (2 	an^4 x - 3 	an^2 x + 6) + C$.

$\int (\tan^7 x + \tan x) dx =$

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