What is the integral of tan^2(t) ?

The integral of tan^2(t) is -t+tan(t)+C

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integral of tan^2(t)

\int tan^{2} (t) d t

- t + tan (t) + C

Solution steps

\int tan^{2} (t) d t

Rewrite using trig identities

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int 1 d t = t

\int sec^{2} (t) d t = tan (t)

= - t + tan (t)

Add a constant to the solution

= - t + tan (t) + C

\int \frac{x ^{3} + 93 x + 70}{x ^{3} + 100 x} d x

y = x^{3} - \frac{x ^{2}}{16} + 4 x + 5

f (x) = 2 x^{2} + 3 x - 2, at x = - 1

\frac{d}{d x} (k (1 - \frac{w}{d} e^{- s x} sin (d x + t)))

- x