What is the integral of sec^4(t/2) ?

The integral of sec^4(t/2) is 2(tan(t/2)+(tan^3(t/2))/3)+C

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integral of sec^4(t/2)

\int sec^{4} (\frac{t}{2}) d t

2 (tan (\frac{t}{2}) + \frac{tan ^{3} ( \frac{t}{2} )}{3}) + C

Solution steps

\int sec^{4} (\frac{t}{2}) d t

Apply u-substitution

: \int sec^{4} (u) \cdot 2 d u

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

Take the constant out:

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

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

Rewrite using trig identities

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

Apply u-substitution

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

= 2 \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

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

\int 1 d v = v

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

= 2 (v + \frac{v ^{3}}{3})

Substitute back

= 2 (tan (\frac{t}{2}) + \frac{tan ^{3} ( \frac{t}{2} )}{3})

Add a constant to the solution

= 2 (tan (\frac{t}{2}) + \frac{tan ^{3} ( \frac{t}{2} )}{3}) + C

\int x^{2} (x - 1)^{2} d x

\frac{2}{( 3 x + 1 )}, x = 2

\int \frac{1}{- x ^{2} + 8 x - 15} d x

\int \frac{e ^{3 x} + 3 e ^{x} + 2}{e ^{x}} d x

f (t) = - \frac{1}{t ^{2}}

What is the integral of sec^4(t/2) ?
The integral of sec^4(t/2) is 2(tan(t/2)+(tan^3(t/2))/3)+C