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

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

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

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

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

Solution steps

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

Simplify

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

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

Apply u-substitution

= \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

= 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{x}{4}

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

Simplify

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

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

Add a constant to the solution

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

t a y l or e^{x} + 0 + 5

x \to 0 lim ((1 + 4 x)^{\frac{3}{x}})

\int (\frac{tan ( u )}{4} - \frac{4}{7} u^{6} - \frac{2}{u}) d u

x \to a lim (\frac{1}{2})

d er i v a t i v e f (x) = (5 x + 4)^{3}

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