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Pattern - 27 of Indefinite Integration

February 19, 2021
Category: Pattern - 27 of Indefinite Integration

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⋆ Pattern – 27

If I_n = ∫ $\tan^n{x}$ dx, then

$I_n=\ \frac{\tan^{n-1\ }{x}}{n-1}-I_n-2\ :n\geq2$

1) $\ I_n=\ \int\tan^6{x}\ dx$

$=\ \int\tan^4{x}\times \tan^2{x}\ dx$

$=\ \int\tan^4{x}\ \left(\sec^2{x}-1\right)\ dx$

$=\ \int\tan^4{x}\times \sec^2{x}\ dx-\ \int\tan^4{x}dx$

$=\ \int\tan^4{x}\times \sec^2{x}\ dx-\ \int\tan^2{x}\times \tan^2{x}\ dx$

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

$=\ \int\tan^4{x}\times \sec^2{x}\ dx-\ \int\tan^2{x}\times \sec^2{x}\ dx+\ \int\tan^2{x}\ dx$

$=\ \int\tan^4{x}\times \sec^2{x}\ dx-\ \int\tan^2{x}\times \sec^2{x}dx+\ \int\left(\sec^2{x}-1\right)dx$

$=\ \frac{\tan^5{x}}{5}-\ \frac{\tan^3{x}}{3}+\tan{x\ }-x+c$

$=\ \frac{\tan^5{x}}{5}-\int\tan^4{x}\ dx$

$=\ \frac{\tan^5{x}}{5}-\left[\frac{\tan^3{x}}{3}-\ \int\tan^2{x}\ dx\right]$

$=\ \frac{\tan^5{x}}{5}-\ \frac{\tan^3{x}}{3}+\left[\tan{x}-\int1dx\right]$

$=\ \frac{\tan^5{x}}{5}-\ \frac{\tan^3{x}}{3}+\tan{x-x+c}$

$=\ \int\tan^n{x}dx$

$=\ \int\tan^{n-2}{x}\times \tan^2{x}\ dx$

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

$=\ \int\tan^{n-2\ }{x}\times \sec^2{x}\ dx-\ \int\tan^{n-2}{x}\ dx$

$=\ \frac{\tan^{n-2\ }{x}}{n-1}-I_{n-2}$

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