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

The integral of tan^{-3/2}(x)sec^4(x) is -2sqrt(cot(x))+2/3 tan^{3/2}(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

- 2 cot (x) + \frac{2}{3} tan^{\frac{3}{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 \frac{1 + u ^{2}}{u ^{\frac{3}{2}}} d u

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

Expand

\frac{1 + u ^{2}}{u ^{\frac{3}{2}}} : \frac{1}{u ^{\frac{3}{2}}} + u^{\frac{1}{2}}

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

\int \frac{1}{u ^{\frac{3}{2}}} d u = - \frac{2}{u}

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

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

Substitute back

u = tan (x)

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

Use the basic trigonometric identity:

\frac{1}{tan ( x )} = cot (x)

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

Add a constant to the solution

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

\frac{d y}{d x} = (1 + y^{2}) tan (x)

\frac{d}{d x} (- 6 e^{- 6 x})

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

\frac{d}{d x} (\frac{( z x - y ) ^{2}}{x})

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