What is the integral of 9xtan^2(3x) ?

The integral of 9xtan^2(3x) is -9x^2+3xtan(3x)+(9x^2)/2+ln|cos(3x)|+C

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integral of 9xtan^2(3x)

\int 9 x tan^{2} (3 x) d x

- 9 x^{2} + 3 x tan (3 x) + \frac{9 x ^{2}}{2} + ln ∣ cos (3 x) ∣ + C

Solution steps

\int 9 x tan^{2} (3 x) d x

Take the constant out:

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

= 9 \cdot \int x tan^{2} (3 x) d x

Rewrite using trig identities

= 9 \cdot \int x (- 1 + sec^{2} (3 x)) d x

Apply u-substitution

: \int \frac{u tan ^{2} ( u )}{9} d u

= 9 \cdot \int \frac{u tan ^{2} ( u )}{9} d u

Take the constant out:

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

= 9 \cdot \frac{1}{9} \cdot \int u tan^{2} (u) d u

Apply Integration By Parts

: u (- u + tan (u)) - \int - u + tan (u) d u

= 9 \cdot \frac{1}{9} (u (- u + tan (u)) - \int - u + tan (u) d u)

\int - u + tan (u) d u = - \frac{u ^{2}}{2} - ln ∣ cos (u) ∣

= 9 \cdot \frac{1}{9} (u (- u + tan (u)) - (- \frac{u ^{2}}{2} - ln ∣ cos (u) ∣))

Substitute back

u = 3 x

= 9 \cdot \frac{1}{9} (3 x (- 3 x + tan (3 x)) - (- \frac{( 3 x ) ^{2}}{2} - ln ∣ cos (3 x) ∣))

Simplify

9 \cdot \frac{1}{9} (3 x (- 3 x + tan (3 x)) - (- \frac{( 3 x ) ^{2}}{2} - ln ∣ cos (3 x) ∣)) : - 9 x^{2} + 3 x tan (3 x) + \frac{9 x ^{2}}{2} + ln ∣ cos (3 x) ∣

= - 9 x^{2} + 3 x tan (3 x) + \frac{9 x ^{2}}{2} + ln ∣ cos (3 x) ∣

Add a constant to the solution

= - 9 x^{2} + 3 x tan (3 x) + \frac{9 x ^{2}}{2} + ln ∣ cos (3 x) ∣ + C

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What is the integral of 9xtan^2(3x) ?
The integral of 9xtan^2(3x) is -9x^2+3xtan(3x)+(9x^2)/2+ln|cos(3x)|+C