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

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

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integral of sqrt(tan(x)sec^4(x))

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

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

Solution steps

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

Apply exponent rule:

(a \cdot b)^{n} = a^{n} \cdot b^{n}

tan (x) sec^{4} (x) = tan (x) sec^{4} (x)

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

sec^{4} (x) = sec^{2} (x),

assuming

sec (x) \geq 0

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

Apply u-substitution

: \int u d u

= \int u d u

Apply the Power Rule

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

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

Substitute back

u = tan (x)

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

Add a constant to the solution

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

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