What is the integral of 17sqrt(tan(x))sec^4(x) ?

The integral of 17sqrt(tan(x))sec^4(x) is 17(2/3 tan^{3/2}(x)+2/7 tan^{7/2}(x))+C

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integral of 17sqrt(tan(x))sec^4(x)

\int 17 tan (x) sec^{4} (x) d x

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

Solution steps

\int 17 tan (x) sec^{4} (x) d x

Take the constant out:

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

= 17 \cdot \int tan (x) sec^{4} (x) d x

Rewrite using trig identities

= 17 \cdot \int (1 + tan^{2} (x)) sec^{2} (x) tan (x) d x

Apply u-substitution

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

= 17 \cdot \int u (1 + u^{2}) d u

Expand

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

= 17 \cdot \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

= 17 (\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}}

= 17 (\frac{2}{3} u^{\frac{3}{2}} + \frac{2}{7} u^{\frac{7}{2}})

Substitute back

u = tan (x)

= 17 (\frac{2}{3} tan^{\frac{3}{2}} (x) + \frac{2}{7} tan^{\frac{7}{2}} (x))

Add a constant to the solution

= 17 (\frac{2}{3} tan^{\frac{3}{2}} (x) + \frac{2}{7} tan^{\frac{7}{2}} (x)) + C

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What is the integral of 17sqrt(tan(x))sec^4(x) ?
The integral of 17sqrt(tan(x))sec^4(x) is 17(2/3 tan^{3/2}(x)+2/7 tan^{7/2}(x))+C