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

The integral of 5xtan^2(3x^2-5)sec^2(3x^2-5) is 5/18 tan^3(3x^2-5)+C

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

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

\frac{5}{18} tan^{3} (3 x^{2} - 5) + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int \frac{tan ^{2} ( u ) sec ^{2} ( u )}{6} d u

= 5 \cdot \int \frac{tan ^{2} ( u ) sec ^{2} ( u )}{6} d u

Take the constant out:

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

= 5 \cdot \frac{1}{6} \cdot \int tan^{2} (u) sec^{2} (u) d u

Rewrite using trig identities

= 5 \cdot \frac{1}{6} \cdot \int (1 + tan^{2} (u)) tan^{2} (u) d u

Apply u-substitution

: \int v^{2} d v

= 5 \cdot \frac{1}{6} \cdot \int v^{2} d v

Apply the Power Rule

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

= 5 \cdot \frac{1}{6} \cdot \frac{v ^{3}}{3}

Substitute back

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

Simplify

5 \cdot \frac{1}{6} \cdot \frac{tan ^{3} ( 3 x ^{2} - 5 )}{3} : \frac{5}{18} tan^{3} (3 x^{2} - 5)

= \frac{5}{18} tan^{3} (3 x^{2} - 5)

Add a constant to the solution

= \frac{5}{18} tan^{3} (3 x^{2} - 5) + C

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What is the integral of 5xtan^2(3x^2-5)sec^2(3x^2-5) ?
The integral of 5xtan^2(3x^2-5)sec^2(3x^2-5) is 5/18 tan^3(3x^2-5)+C