What is the integral of tan^8(x/3)sec^2(x/3) ?

The integral of tan^8(x/3)sec^2(x/3) is 1/3 tan^9(x/3)+C

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integral of tan^8(x/3)sec^2(x/3)

\int tan^{8} (\frac{x}{3}) sec^{2} (\frac{x}{3}) d x

\frac{1}{3} tan^{9} (\frac{x}{3}) + C

Solution steps

\int tan^{8} (\frac{x}{3}) sec^{2} (\frac{x}{3}) d x

Apply u-substitution

: \int tan^{8} (u) sec^{2} (u) \cdot 3 d u

= \int tan^{8} (u) sec^{2} (u) \cdot 3 d u

Take the constant out:

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

= 3 \cdot \int tan^{8} (u) sec^{2} (u) d u

Rewrite using trig identities

= 3 \cdot \int (1 + tan^{2} (u)) tan^{8} (u) d u

Apply u-substitution

: \int v^{8} d v

= 3 \cdot \int v^{8} d v

Apply the Power Rule

: \frac{v ^{9}}{9}

= 3 \cdot \frac{v ^{9}}{9}

Substitute back

= 3 \cdot \frac{tan ^{9} ( \frac{x}{3} )}{9}

Simplify

3 \cdot \frac{tan ^{9} ( \frac{x}{3} )}{9} : \frac{1}{3} tan^{9} (\frac{x}{3})

= \frac{1}{3} tan^{9} (\frac{x}{3})

Add a constant to the solution

= \frac{1}{3} tan^{9} (\frac{x}{3}) + C

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What is the integral of tan^8(x/3)sec^2(x/3) ?
The integral of tan^8(x/3)sec^2(x/3) is 1/3 tan^9(x/3)+C