What is the integral of 1/(cosh^2(x)) ?

The integral of 1/(cosh^2(x)) is tanh(x)+C

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integral of 1/(cosh^2(x))

\int \frac{1}{cosh ^{2} ( x )} d x

tanh (x) + C

Solution steps

\int \frac{1}{cosh ^{2} ( x )} d x

Use the Hyperbolic identity:

\frac{1}{cosh ( x )} = sech (x)

= \int sech^{2} (x) d x

Use the common integral:

\int sech^{2} (x) d x = tanh (x)

= tanh (x)

Add a constant to the solution

= tanh (x) + C

x \to \frac{π}{2} + lim (\frac{1}{cos ( x )})

f (t) = t e^{- 3 t} cos (5 t)

\frac{d}{d x} (v) = \frac{1}{2 v}

\int (4 - x) 8^{(4 - x)^{2}} d x

y^{^{''}} - 4 y^{^{'}} + 4 y = e^{- t}, y (0) = 0, y^{^{'}} (0) = 0

What is the integral of 1/(cosh^2(x)) ?
The integral of 1/(cosh^2(x)) is tanh(x)+C