What is the integral of 1/(sec^4(x)) ?

The integral of 1/(sec^4(x)) is cos^3(x)sin(x)+3/8 (x-1/4 sin(4x))+C

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integral of 1/(sec^4(x))

\int \frac{1}{sec ^{4} ( x )} d x

cos^{3} (x) sin (x) + \frac{3}{8} (x - \frac{1}{4} sin (4 x)) + C

Solution steps

\int \frac{1}{sec ^{4} ( x )} d x

Rewrite using trig identities

= \int cos^{4} (x) d x

Simplify

cos^{4} (x) : cos^{3} (x) cos (x)

= \int cos^{3} (x) cos (x) d x

Apply Integration By Parts

: cos^{3} (x) sin (x) - \int - 3 cos^{2} (x) sin^{2} (x) d x

= cos^{3} (x) sin (x) - \int - 3 cos^{2} (x) sin^{2} (x) d x

\int - 3 cos^{2} (x) sin^{2} (x) d x = - \frac{3}{8} (x - \frac{1}{4} sin (4 x))

= cos^{3} (x) sin (x) - (- \frac{3}{8} (x - \frac{1}{4} sin (4 x)))

Simplify

= cos^{3} (x) sin (x) + \frac{3}{8} (x - \frac{1}{4} sin (4 x))

Add a constant to the solution

= cos^{3} (x) sin (x) + \frac{3}{8} (x - \frac{1}{4} sin (4 x)) + C

- 2 e^{2 t}

\frac{s - 5}{( s - 4 ) ^{2} - 3}

\int_{1}^{2} \frac{8 - x ^{2}}{x ^{3}} d x

\int \frac{x ^{2}}{2} + 1 d x

\frac{\partial}{\partial x} (x y + 81)

What is the integral of 1/(sec^4(x)) ?
The integral of 1/(sec^4(x)) is cos^3(x)sin(x)+3/8 (x-1/4 sin(4x))+C