What is the integral of-5x^2sec^2(3x^3)tan^5(3x^3) ?

The integral of-5x^2sec^2(3x^3)tan^5(3x^3) is -5/54 (sec^6(3x^3)-3sec^4(3x^3)+3sec^2(3x^3))+C

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integral of-5x^2sec^2(3x^3)tan^5(3x^3)

\int - 5 x^{2} sec^{2} (3 x^{3}) tan^{5} (3 x^{3}) d x

- \frac{5}{54} (sec^{6} (3 x^{3}) - 3 sec^{4} (3 x^{3}) + 3 sec^{2} (3 x^{3})) + C

Solution steps

\int - 5 x^{2} sec^{2} (3 x^{3}) tan^{5} (3 x^{3}) d x

Take the constant out:

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

= - 5 \cdot \int x^{2} sec^{2} (3 x^{3}) tan^{5} (3 x^{3}) d x

Apply u-substitution

: \int \frac{u tan ^{5} ( 3 u ^{\frac{3}{2}} ) sec ^{2} ( 3 u ^{\frac{3}{2}} )}{2 x} d u

= - 5 \cdot \int \frac{tan ^{5} ( 3 u ) sec ^{2} ( 3 u )}{3} d u

Take the constant out:

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

= - 5 \cdot \frac{1}{3} \cdot \int tan^{5} (3 u) sec^{2} (3 u) d u

Apply u-substitution

: \int tan^{5} (v) sec^{2} (v) \frac{1}{3} d v

= - 5 \cdot \frac{1}{3} \cdot \int tan^{5} (v) sec^{2} (v) \frac{1}{3} d v

Take the constant out:

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

= - 5 \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \int tan^{5} (v) sec^{2} (v) d v

Rewrite using trig identities

= - 5 \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \int (- 1 + sec^{2} (v))^{2} tan (v) sec^{2} (v) d v

Apply u-substitution

: \int w (w^{2} - 1)^{2} d w

= - 5 \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \int w (w^{2} - 1)^{2} d w

Expand

w (w^{2} - 1)^{2} : w^{5} - 2 w^{3} + w

= - 5 \cdot \frac{1}{3} \cdot \frac{1}{3} \cdot \int w^{5} - 2 w^{3} + w d w

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - 5 \cdot \frac{1}{3} \cdot \frac{1}{3} (\int w^{5} d w - \int 2 w^{3} d w + \int w d w)

\int w^{5} d w = \frac{w ^{6}}{6}

\int 2 w^{3} d w = \frac{w ^{4}}{2}

\int w d w = \frac{w ^{2}}{2}

= - 5 \cdot \frac{1}{3} \cdot \frac{1}{3} (\frac{w ^{6}}{6} - \frac{w ^{4}}{2} + \frac{w ^{2}}{2})

Substitute back

= - 5 \cdot \frac{1}{3} \cdot \frac{1}{3} (\frac{sec ^{6} ( 3 x ^{3} )}{6} - \frac{sec ^{4} ( 3 x ^{3} )}{2} + \frac{sec ^{2} ( 3 x ^{3} )}{2})

Simplify

- 5 \cdot \frac{1}{3} \cdot \frac{1}{3} (\frac{sec ^{6} ( 3 x ^{3} )}{6} - \frac{sec ^{4} ( 3 x ^{3} )}{2} + \frac{sec ^{2} ( 3 x ^{3} )}{2}) : - \frac{5}{54} (sec^{6} (3 x^{3}) - 3 sec^{4} (3 x^{3}) + 3 sec^{2} (3 x^{3}))

= - \frac{5}{54} (sec^{6} (3 x^{3}) - 3 sec^{4} (3 x^{3}) + 3 sec^{2} (3 x^{3}))

Add a constant to the solution

= - \frac{5}{54} (sec^{6} (3 x^{3}) - 3 sec^{4} (3 x^{3}) + 3 sec^{2} (3 x^{3})) + C

x^{3} - x, x^{2} + x, - 1, 2

\frac{6 x}{x ^{2} + 4}

\frac{d y}{d x} = x^{2} y^{2}

f (x) = (ln (x))^{5}, at x = 4

x \to 0 lim (1 + cos (x))

What is the integral of-5x^2sec^2(3x^3)tan^5(3x^3) ?
The integral of-5x^2sec^2(3x^3)tan^5(3x^3) is -5/54 (sec^6(3x^3)-3sec^4(3x^3)+3sec^2(3x^3))+C