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

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

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integral of sec(3-x)tan(3-x)

\int sec (3 - x) tan (3 - x) d x

- sec (3 - x) + C

Solution steps

\int sec (3 - x) tan (3 - x) d x

Apply u-substitution

: \int - sec (u) tan (u) d u

= \int - sec (u) tan (u) d u

Take the constant out:

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

= - \int sec (u) tan (u) d u

Apply u-substitution

: \int 1 d v

= - \int 1 d v

Integral of a constant:

\int a d x = a x

= - 1 \cdot v

Substitute back

= - 1 \cdot sec (3 - x)

Simplify

= - sec (3 - x)

Add a constant to the solution

= - sec (3 - x) + C

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

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

\int (y - \frac{sin ^{2} ( x )}{y ^{2}}) d y

\int \frac{1}{z ln ( z )} d z

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

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