What is the integral of 5sec(3x^2+4)tan(3x^2+4)x ?

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

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integral of 5sec(3x^2+4)tan(3x^2+4)x

\int 5 sec (3 x^{2} + 4) tan (3 x^{2} + 4) x d x

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

Solution steps

\int 5 sec (3 x^{2} + 4) tan (3 x^{2} + 4) x d x

Take the constant out:

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

= 5 \cdot \int sec (3 x^{2} + 4) tan (3 x^{2} + 4) x d x

Apply u-substitution

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

= 5 \cdot \int \frac{sec ( u ) tan ( 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 sec (u) tan (u) d u

Apply u-substitution

: \int 1 d v

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

Integral of a constant:

\int a d x = a x

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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