What is the integral of-sin(x)sec(x)tan(x) ?

The integral of-sin(x)sec(x)tan(x) is -tan(x)+x+C

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integral of-sin(x)sec(x)tan(x)

\int - sin (x) sec (x) tan (x) d x

- tan (x) + x + C

Solution steps

\int - sin (x) sec (x) tan (x) d x

Take the constant out:

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

= - \int sin (x) sec (x) tan (x) d x

Rewrite using trig identities

= - \int \frac{1 - cos ^{2} ( x )}{cos ^{2} ( x )} d x

Expand

\frac{1 - cos ^{2} ( x )}{cos ^{2} ( x )} : \frac{1}{cos ^{2} ( x )} - 1

= - \int \frac{1}{cos ^{2} ( x )} - 1 d x

Apply the Sum Rule:

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

= - (\int \frac{1}{cos ^{2} ( x )} d x - \int 1 d x)

\int \frac{1}{cos ^{2} ( x )} d x = tan (x)

\int 1 d x = x

= - (tan (x) - x)

Simplify

= - tan (x) + x

Add a constant to the solution

= - tan (x) + x + C

\int 4 x^{3} - \frac{3}{x ^{4}} d x

3 - 9 x

\frac{d}{d x} (4 x - \frac{3}{x} + \frac{1}{x ^{2}})

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

\int \frac{1}{2 x} cos (2 x) d x

What is the integral of-sin(x)sec(x)tan(x) ?
The integral of-sin(x)sec(x)tan(x) is -tan(x)+x+C