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

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

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integral of tan^2(y)

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

- y + tan (y) + C

Solution steps

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

Rewrite using trig identities

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

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 y + \int sec^{2} (y) d y

\int 1 d y = y

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

= - y + tan (y)

Add a constant to the solution

= - y + tan (y) + C

y = \frac{1}{x ^{3}}

\int_{- 2}^{2} x^{3} - 1 d x

z (v) = v^{\frac{3}{2}} (v + c e^{v})

y^{^{''}} + y = (cot (x))^{2}

\frac{\partial}{\partial x} (\frac{6 x}{x ^{2} + 5})