What is the integral of 1/(sec^3(y)) ?

The integral of 1/(sec^3(y)) is sin(y)-(sin^3(y))/3+C

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integral of 1/(sec^3(y))

\int \frac{1}{sec ^{3} ( y )} d y

sin (y) - \frac{sin ^{3} ( y )}{3} + C

Solution steps

\int \frac{1}{sec ^{3} ( y )} d y

Rewrite using trig identities

= \int (1 - sin^{2} (y)) cos (y) d y

Apply u-substitution

: \int 1 - u^{2} d u

= \int 1 - u^{2} d u

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 u - \int u^{2} d u

\int 1 d u = u

\int u^{2} d u = \frac{u ^{3}}{3}

= u - \frac{u ^{3}}{3}

Substitute back

u = sin (y)

= sin (y) - \frac{sin ^{3} ( y )}{3}

Add a constant to the solution

= sin (y) - \frac{sin ^{3} ( y )}{3} + C

f (θ) = θ cos (θ)

\int_{0}^{6} x^{2} d x

sec (x) - 22 tan (x)

\frac{\partial}{\partial x} (4 y^{2} e^{7 x})

y = \frac{8 e ^{8 x}}{7 x + 2}

What is the integral of 1/(sec^3(y)) ?
The integral of 1/(sec^3(y)) is sin(y)-(sin^3(y))/3+C