What is the integral of tan^{3/2}(x)sec^4(x) ?

The integral of tan^{3/2}(x)sec^4(x) is 2/5 tan^{5/2}(x)+2/9 tan^{9/2}(x)+C

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integral of tan^{3/2}(x)sec^4(x)

\int tan^{\frac{3}{2}} (x) sec^{4} (x) d x

\frac{2}{5} tan^{\frac{5}{2}} (x) + \frac{2}{9} tan^{\frac{9}{2}} (x) + C

Solution steps

\int tan^{\frac{3}{2}} (x) sec^{4} (x) d x

Rewrite using trig identities

= \int (1 + tan^{2} (x)) sec^{2} (x) tan^{\frac{3}{2}} (x) d x

Apply u-substitution

: \int u^{\frac{3}{2}} (1 + u^{2}) d u

= \int u^{\frac{3}{2}} (1 + u^{2}) d u

Expand

u^{\frac{3}{2}} (1 + u^{2}) : u^{\frac{3}{2}} + u^{\frac{7}{2}}

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

\int u^{\frac{3}{2}} d u = \frac{2}{5} u^{\frac{5}{2}}

\int u^{\frac{7}{2}} d u = \frac{2}{9} u^{\frac{9}{2}}

= \frac{2}{5} u^{\frac{5}{2}} + \frac{2}{9} u^{\frac{9}{2}}

Substitute back

u = tan (x)

= \frac{2}{5} tan^{\frac{5}{2}} (x) + \frac{2}{9} tan^{\frac{9}{2}} (x)

Add a constant to the solution

= \frac{2}{5} tan^{\frac{5}{2}} (x) + \frac{2}{9} tan^{\frac{9}{2}} (x) + C

\int \frac{3 x + 16}{x ^{2} - x - 6} d x

x \to 1 lim (0.8 x + 1.2)

x \to 0 lim (cos (x) sin (x))

y^{^{'}} = 2 y + 3

\frac{d x}{d t} = (a - b \cdot t) x

What is the integral of tan^{3/2}(x)sec^4(x) ?
The integral of tan^{3/2}(x)sec^4(x) is 2/5 tan^{5/2}(x)+2/9 tan^{9/2}(x)+C