If int tan ^4 x dx = a tan ^3 x + b tan x + c | vedclass

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Question

(A) We have $I = \int 	an^4 x dx = \int 	an^2 x (\sec^2 x - 1) dx$. $I = \int 	an^2 x \sec^2 x dx - \int 	an^2 x dx$. $I = \int 	an^2 x \sec^2 x

Vedclass · Accepted Answer

$\frac{7}{3}$

Vedclass · Answer

(A) We have $I = \int 	an^4 x dx = \int 	an^2 x (\sec^2 x - 1) dx$. $I = \int 	an^2 x \sec^2 x dx - \int 	an^2 x dx$. $I = \int 	an^2 x \sec^2 x dx - \int (\sec^2 x - 1) dx$. $I = \frac{	an^3 x}{3} - 	an x + x + k$. Comparing this with $a 	an^3 x + b 	an x + c x + k$,we get $a = \frac{1}{3}$,$b = -1$,and $c = 1$. Therefore,$a - b + c = \frac{1}{3} - (-1) + 1 = \frac{1}{3} + 1 + 1 = \frac{1}{3} + 2 = \frac{7}{3}$.

If $\int \tan ^4 x dx = a \tan ^3 x + b \tan x + c x + k$ (where $k$ is the constant of integration),then the value of $a - b + c =$

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