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

The integral of 1/(sec^4(1/4 β)) is 4(cos^3(β/4)sin(β/4)+3/32 (β-sin(β)))+C

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integral of 1/(sec^4(1/4 β))

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

4 (cos^{3} (\frac{β}{4}) sin (\frac{β}{4}) + \frac{3}{32} (β - sin (β))) + C

Solution steps

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

Simplify

\frac{1}{sec ^{4} ( \frac{1}{4} β )} : cos^{4} (\frac{β}{4})

= \int cos^{4} (\frac{β}{4}) d β

Apply u-substitution

: \int cos^{4} (u) \cdot 4 d u

= \int cos^{4} (u) \cdot 4 d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int cos^{4} (u) d u

Simplify

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

= 4 \cdot \int cos^{3} (u) cos (u) d u

Apply Integration By Parts

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

= 4 (cos^{3} (u) sin (u) - \int - 3 cos^{2} (u) sin^{2} (u) d u)

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

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

Substitute back

u = \frac{β}{4}

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

Simplify

4 (cos^{3} (\frac{β}{4}) sin (\frac{β}{4}) - (- \frac{3}{8} (\frac{β}{4} - \frac{1}{4} sin (4 \cdot \frac{β}{4})))) : 4 (cos^{3} (\frac{β}{4}) sin (\frac{β}{4}) + \frac{3}{32} (β - sin (β)))

= 4 (cos^{3} (\frac{β}{4}) sin (\frac{β}{4}) + \frac{3}{32} (β - sin (β)))

Add a constant to the solution

= 4 (cos^{3} (\frac{β}{4}) sin (\frac{β}{4}) + \frac{3}{32} (β - sin (β))) + C

\int 6 \cdot x + 4 d x

\frac{\partial}{\partial x} (sin (x + 2 y + 3 z))

f (x) = \frac{4 x}{3 x ^{2} + 1}

x \to 4 lim (\frac{7}{( x - 4 )})

\frac{\partial}{\partial x} (y ln (x y))

What is the integral of 1/(sec^4(1/4 β)) ?
The integral of 1/(sec^4(1/4 β)) is 4(cos^3(β/4)sin(β/4)+3/32 (β-sin(β)))+C