What is the integral of tan^2(1/3 x)sec^2(1/3 x) ?

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

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integral of tan^2(1/3 x)sec^2(1/3 x)

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

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

Solution steps

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

Rewrite using trig identities

= \int \frac{sin ^{2} ( \frac{x}{3} ) sec ^{2} ( \frac{x}{3} )}{cos ^{2} ( \frac{x}{3} )} d x

Simplify

\frac{sin ^{2} ( \frac{x}{3} ) sec ^{2} ( \frac{x}{3} )}{cos ^{2} ( \frac{x}{3} )} : sec^{2} (\frac{x}{3}) tan^{2} (\frac{x}{3})

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

Apply u-substitution

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

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

Rewrite using trig identities

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

Apply u-substitution

: \int v^{2} d v

= 3 \cdot \int v^{2} d v

Apply the Power Rule

: \frac{v ^{3}}{3}

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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