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Q.Obtain reduction formula for In=∫tan⁡nx dxI_n = \int \tan^n x\, dx, for an integer n≥2n \geq 2; deduce the value of ∫tan⁡6x dx\int \tan^6 x\, dx.

Telangana TsbieTelangana Board of Intermediate Education 2025Subjective· 7mImportance★★★★★
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In=∫tan⁡n−2x(sec⁡2x−1) dx=tan⁡n−1xn−1−In−2I_n=\int\tan^{n-2}x(\sec^2x-1)\,dx=\dfrac{\tan^{n-1}x}{n-1}-I_{n-2}; unwinding for n=6n=6 gives tan⁡5x5−tan⁡3x3+tan⁡x−x+C\tfrac{\tan^5x}{5}-\tfrac{\tan^3x}{3}+\tan x-x+C.

Deriving the reduction formula:

In=∫tan⁡nx dx=∫tan⁡n−2x tan⁡2x dx=∫tan⁡n−2x (sec⁡2x−1) dx.I_n=\int\tan^n x\,dx=\int\tan^{n-2}x\,\tan^2 x\,dx=\int\tan^{n-2}x\,(\sec^2 x-1)\,dx.

=∫tan⁡n−2x sec⁡2x dx−∫tan⁡n−2x dx.=\int\tan^{n-2}x\,\sec^2 x\,dx-\int\tan^{n-2}x\,dx.

The first integral, with u=tan⁡xu=\tan x, is tan⁡n−1xn−1\dfrac{\tan^{n-1}x}{n-1}. Hence

In=tan⁡n−1xn−1−In−2.I_n=\dfrac{\tan^{n-1}x}{n-1}-I_{n-2}.

Apply for n=6n=6:

I6=tan⁡5x5−I4,I4=tan⁡3x3−I2,I2=tan⁡x1−I0.I_6=\dfrac{\tan^5x}{5}-I_4,\quad I_4=\dfrac{\tan^3x}{3}-I_2,\quad I_2=\dfrac{\tan x}{1}-I_0.

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