What is the integral of tan^{1/2}(x)sec^4(x) ?

The integral of tan^{1/2}(x)sec^4(x) is 2/3 tan^{3/2}(x)+2/7 tan^{7/2}(x)+C

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integral of tan^{1/2}(x)sec^4(x)

\int tan^{\frac{1}{2}} (x) sec^{4} (x) d x

\frac{2}{3} tan^{\frac{3}{2}} (x) + \frac{2}{7} tan^{\frac{7}{2}} (x) + C

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

: \int u (1 + u^{2}) d u

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

Expand

u (1 + u^{2}) : u + u^{\frac{5}{2}}

= \int u + u^{\frac{5}{2}} d u

Apply the Sum Rule:

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

= \int u d u + \int u^{\frac{5}{2}} d u

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

\int u^{\frac{5}{2}} d u = \frac{2}{7} u^{\frac{7}{2}}

= \frac{2}{3} u^{\frac{3}{2}} + \frac{2}{7} u^{\frac{7}{2}}

Substitute back

u = tan (x)

= \frac{2}{3} tan^{\frac{3}{2}} (x) + \frac{2}{7} tan^{\frac{7}{2}} (x)

Add a constant to the solution

= \frac{2}{3} tan^{\frac{3}{2}} (x) + \frac{2}{7} tan^{\frac{7}{2}} (x) + C

\frac{d}{d x} (\frac{1}{3} x^{3} + x^{2})

\frac{( s + 1 )}{s ^{2} + s + \frac{1}{2}}

\int \frac{6}{( x - 2 ) ( x + 4 )} d x

\frac{\partial}{\partial x} ((3 x + 4 y)^{5})

y = \frac{x ^{5}}{1 - x ^{4}}

What is the integral of tan^{1/2}(x)sec^4(x) ?
The integral of tan^{1/2}(x)sec^4(x) is 2/3 tan^{3/2}(x)+2/7 tan^{7/2}(x)+C