What is the integral of tan^2(y)+1 ?

The integral of tan^2(y)+1 is tan(y)+C

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integral of tan^2(y)+1

\int tan^{2} (y) + 1 d y

tan (y) + C

Solution steps

\int tan^{2} (y) + 1 d y

Rewrite using trig identities

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

Use the common integral:

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

= tan (y)

Add a constant to the solution

= tan (y) + C

\frac{\partial}{\partial x} (2 x^{2} + y^{2} - 4 y + 4)

0.5 (e^{x} + e^{- x}), 0 \leq x \leq ln (2)

\frac{d}{d x} (\frac{lo g _{10} ( x )}{ln ( x )} + 2 x)

y^{^{''}} + 5 y^{^{'}} + 6 y = 2 e^{x}

\frac{d}{d x} (in (sec (x)))