What is the integral of (tan(x)(6cos(x)-6sec(x))) ?

The integral of (tan(x)(6cos(x)-6sec(x))) is -6cos(x)-6sec(x)+C

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integral of (tan(x)(6cos(x)-6sec(x)))

\int (tan (x) (6 cos (x) - 6 sec (x))) d x

- 6 cos (x) - 6 sec (x) + C

Solution steps

\int tan (x) (6 cos (x) - 6 sec (x)) d x

Rewrite using trig identities

= \int \frac{sin ( x ) ( 6 cos ^{2} ( x ) - 6 )}{cos ^{2} ( x )} d x

Expand

\frac{sin ( x ) ( 6 cos ^{2} ( x ) - 6 )}{cos ^{2} ( x )} : 6 sin (x) - \frac{6 sin ( x )}{cos ^{2} ( x )}

= \int 6 sin (x) - \frac{6 sin ( x )}{cos ^{2} ( x )} 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 6 sin (x) d x - \int \frac{6 sin ( x )}{cos ^{2} ( x )} d x

\int 6 sin (x) d x = - 6 cos (x)

\int \frac{6 sin ( x )}{cos ^{2} ( x )} d x = 6 sec (x)

= - 6 cos (x) - 6 sec (x)

Add a constant to the solution

= - 6 cos (x) - 6 sec (x) + C

y^{^{''}} - y^{^{'}} - 6 y = - 4 e^{2} t, y (0) = 0, y^{^{'}} (0) = 0

\int \frac{x ^{2}}{( x - 1 ) ( x ^{2} + 4 x + 5 )} d x

\frac{d y}{d x} = \frac{5 x}{3 y}

\int_{0}^{π} 13 sin^{2} (2 x) d x

\int_{1}^{9} t ln (t) d t

What is the integral of (tan(x)(6cos(x)-6sec(x))) ?
The integral of (tan(x)(6cos(x)-6sec(x))) is -6cos(x)-6sec(x)+C