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If $I_{n} = \int \tan^{n} xdx$ then $(n - 1) (I_{n} + I_{n - 2})$ is equal to

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detailed solution

Correct option is C

Here,In=∫tann⁡xdx=∫tann−2⁡xtan2⁡xdx=∫tann−2⁡sec2⁡x−1dx=∫tann−2⁡x⋅sec2⁡xdx−∫tann−2⁡xdx=∫tann−2⁡xsec2⁡xdx−In−2In+In−2=tann−1⁡xn−1Hence, (n−1)In+In−2=tann−1⁡x

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If In=∫tann⁡xdx then (n−1)In+In−2 is equal to

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If $I_{n} = \int \tan^{n} xdx$ then $(n - 1) (I_{n} + I_{n - 2})$ is equal to