What is the integral of 1/3 sec(x)tan(x) ?

The integral of 1/3 sec(x)tan(x) is 1/3 sec(x)+C

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integral of 1/3 sec(x)tan(x)

\int \frac{1}{3} sec (x) tan (x) d x

\frac{1}{3} sec (x) + C

Solution steps

\int \frac{1}{3} sec (x) tan (x) d x

Take the constant out:

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

= \frac{1}{3} \cdot \int sec (x) tan (x) d x

Apply u-substitution

: \int 1 d u

= \frac{1}{3} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= \frac{1}{3} \cdot 1 \cdot u

Substitute back

u = sec (x)

= \frac{1}{3} \cdot 1 \cdot sec (x)

Simplify

= \frac{1}{3} sec (x)

Add a constant to the solution

= \frac{1}{3} sec (x) + C

\frac{d}{d x} (- \frac{1}{x ^{2}} + 1)

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y = (3 x^{2} - 4 x) e^{5 x}

\int 2 sin (3 x) cos (3 x) d x

\int \frac{1}{9 - x ^{2}} d x

What is the integral of 1/3 sec(x)tan(x) ?
The integral of 1/3 sec(x)tan(x) is 1/3 sec(x)+C