Solve : $\int {tan}^{2} x d x$

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Solution

Given,

$\int {tan}^{2} (x) d x$

$= \int - 1 + {sec}^{2} (x) d x$

$= - \int 1 d x + \int {sec}^{2} (x) d x$

$= - x + tan (x) + C$

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Similar questions

I = \int \frac{(1 + {tan}^{2} x)^{2}}{1 + {tan}^{2} x} {sec}^{2} x d x =

\int_{0}^{π / 4} {tan}^{2} x d x =

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I_{1} = \int_{{sec}^{2} z}^{2 - {tan}^{2} z} x f (x (3 - x)) d x

I_{2} = \int_{{sec}^{2} z}^{2 - {tan}^{2} z} f (x (3 - x)) d x

f

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\frac{I_{1}}{I_{2}} =

tan 1^{\circ} tan 2^{\circ} . . . tan 89^{\circ} =

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