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

The integral of sec^4(x/4) is 4(tan(x/4)+(tan^3(x/4))/3)+C

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integral of sec^4(x/4)

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

4 (tan (\frac{x}{4}) + \frac{tan ^{3} ( \frac{x}{4} )}{3}) + C

Solution steps

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

Apply u-substitution

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

= \int sec^{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 sec^{4} (u) d u

Rewrite using trig identities

= 4 \cdot \int (1 + tan^{2} (u)) sec^{2} (u) d u

Apply u-substitution

: \int 1 + v^{2} d v

= 4 \cdot \int 1 + v^{2} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 4 (\int 1 d v + \int v^{2} d v)

\int 1 d v = v

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

= 4 (v + \frac{v ^{3}}{3})

Substitute back

= 4 (tan (\frac{x}{4}) + \frac{tan ^{3} ( \frac{x}{4} )}{3})

Add a constant to the solution

= 4 (tan (\frac{x}{4}) + \frac{tan ^{3} ( \frac{x}{4} )}{3}) + C

\frac{d}{d x} (\frac{5}{( x - 6 ) ^{2}})

\int_{\frac{1}{5}}^{3} 8 x ln (5 x) d x

ln (e)

\int x^{\frac{3}{4}} d x

y^{^{''}} + 289 y = sec (17 x)

What is the integral of sec^4(x/4) ?
The integral of sec^4(x/4) is 4(tan(x/4)+(tan^3(x/4))/3)+C