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Pattern - 29 of Indefinite Integration

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⋆ Pattern – 29

= ∫ \sec^n{x} dx , then

I_n=\ \frac{\tan{x\times \sec^{n-2}{x}}}{n-1}+\frac{n-2}{n-1}\ I_{n-2}\ :n\geq2

Here,  = ∫\sec^n{x}\ dx ——-(1)

\therefore\ \ I_n=\ \int\sec^{n-2\ }{x}\times\sec^2{x}dx

=\sec^{n-2}{x}\ \int\sec^2{x}\ dx-\ \int\left(\frac{d}{dx}\sec^{n-2}{x}\ \int\sec^2{x}\right)dx

=\sec^{n-2}{x} \times tan x -\ \int\left(n-2\right)\sec^{n-2\ }{x} \times sec \times tan x \times tan \times dx

=\sec^{n-2\ }{x} \times tan x-(n - 2) \int \sec^{n-2\ }{x}\ \left(\sec^2{x}-1\right)\ dx

= \sec^{n-2\ }{x}\times tan x- (n-2) ∫ \sec^n{x} dx+ ( n - 2) ∫ \sec^{n-2\ }{x}\ dx

= \sec^{n-2\ }{x}\times tan x- (n-2) I_n + ( n - 2 ) I_{n-2}

Here,

I_n=\sec^{n-2}{x}\ tan x - (n-2)\ I_n + ( n - 2 ) I_{n-2}

∴  [1+(n-2)] I_n = \sec^{n-2}{x} tan x + (n-2) I_{n-2}

∴  (n - 1) I_n = \sec^{n-2}{x} tan x + (n-2) I_{n-2}

\therefore\ \ I_n =  \ \frac{\sec^{n-2\ }{x}\tan{x}}{n-1\ }+\ \frac{n-2}{n-1\ }I_{n-2}

1) Find ∫ \sec^6{x} dx

Solution. Let, I_6 = ∫ \sec^6{x} dx

Here,

I_6=\ \frac{\tan{x}\times \sec^4{x}}{5}+\frac{4}{5}\ I_4

=\ \frac{\tan{x}\times \sec^4{x}}{5}+\frac{4}{5}\ \left[\frac{\tan{x}\times \sec^2{x}}{3}+\frac{2}{3}\ I_2\right]

=\ \frac{\tan{x}\times \sec^4{x}}{5}+\frac{4}{15}\tan{x}\times \sec^2{x}+\frac{8}{15}\int\sec^2{x}\ dx

=\ \frac{\tan{x}\times \sec^4{x}}{5}+\ \frac{4\tan{x}\times \sec^2{x}}{15}+\ \frac{8\tan{x}}{15}+c

 

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