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

The integral of 3x^2sec^2(3x)tan(3x) is 1/18 (9x^2tan^2(3x)-6xtan(3x)-2ln|cos(3x)|+9x^2)+C

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

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

\frac{1}{18} (9 x^{2} tan^{2} (3 x) - 6 x tan (3 x) - 2 ln ∣ cos (3 x) ∣ + 9 x^{2}) + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{u ^{2} sec ^{2} ( u ) tan ( u )}{27} d u

= 3 \cdot \int \frac{u ^{2} sec ^{2} ( u ) tan ( u )}{27} d u

Take the constant out:

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

= 3 \cdot \frac{1}{27} \cdot \int u^{2} sec^{2} (u) tan (u) d u

Apply Integration By Parts

: \frac{1}{2} u^{2} tan^{2} (u) - \int u tan^{2} (u) d u

= 3 \cdot \frac{1}{27} (\frac{1}{2} u^{2} tan^{2} (u) - \int u tan^{2} (u) d u)

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

= 3 \cdot \frac{1}{27} (\frac{1}{2} u^{2} tan^{2} (u) - (- u^{2} + u tan (u) + \frac{u ^{2}}{2} + ln ∣ cos (u) ∣))

Substitute back

u = 3 x

= 3 \cdot \frac{1}{27} (\frac{1}{2} (3 x)^{2} tan^{2} (3 x) - (- (3 x)^{2} + 3 x tan (3 x) + \frac{( 3 x ) ^{2}}{2} + ln ∣ cos (3 x) ∣))

Simplify

3 \cdot \frac{1}{27} (\frac{1}{2} (3 x)^{2} tan^{2} (3 x) - (- (3 x)^{2} + 3 x tan (3 x) + \frac{( 3 x ) ^{2}}{2} + ln ∣ cos (3 x) ∣)) : \frac{1}{18} (9 x^{2} tan^{2} (3 x) - 6 x tan (3 x) - 2 ln ∣ cos (3 x) ∣ + 9 x^{2})

= \frac{1}{18} (9 x^{2} tan^{2} (3 x) - 6 x tan (3 x) - 2 ln ∣ cos (3 x) ∣ + 9 x^{2})

Add a constant to the solution

= \frac{1}{18} (9 x^{2} tan^{2} (3 x) - 6 x tan (3 x) - 2 ln ∣ cos (3 x) ∣ + 9 x^{2}) + C

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What is the integral of 3x^2sec^2(3x)tan(3x) ?
The integral of 3x^2sec^2(3x)tan(3x) is 1/18 (9x^2tan^2(3x)-6xtan(3x)-2ln|cos(3x)|+9x^2)+C