lf $\int T a n^{4} x d x = A {tan}^{3} x + B tan x + C x + D$ , then the descending order of $A, B, C$ is

A, B, C

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B, C, A

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C, A, B

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C, B, A

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Solution

The correct option is B C, A, B
$\int T a n^{4} x d x = \int T a n^{2} x (s e c^{2} x - 1) d x$
$\int T a n^{2} x s e c^{2} x d x - \int T a n^{2} x d x$
$= \int T a n^{2} x d (t a n x) - \int T a n^{2} x d x$
$= \frac{T a n^{3} x}{3} + c - \int [s e c^{2} x - 1] d x$
$= \frac{T a n^{3} x}{3} + c + [x] - T a n x$
$= \frac{T a n^{3} x}{3} - T a n x + x + c$
$A = \frac{1}{3}; B = - 1; C = 1$
So, descending order
C, A, B

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lf ∫Tan4xdx=Atan3x+Btanx+Cx+D, then the descending order of A, B, C is

The correct option is B C, A, B∫Tan4xdx=∫Tan2x(sec2x−1)dx∫Tan2xsec2xdx−∫Tan2xdx=∫Tan2xd(tanx)−∫Tan2xdx=Tan3x3+c−∫[sec2x−1]dx=Tan3x3+c+[x]−Tanx=Tan3x3−Tanx+x+cA=13;B=−1;C=1So, descending orderC, A, B

lf $\int T a n^{4} x d x = A {tan}^{3} x + B tan x + C x + D$ , then the descending order of $A, B, C$ is