What is the integral of 8sec(x)(sec(x)-4tan(x)) ?

The integral of 8sec(x)(sec(x)-4tan(x)) is 8(tan(x)-4sec(x))+C

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integral of 8sec(x)(sec(x)-4tan(x))

\int 8 sec (x) (sec (x) - 4 tan (x)) d x

8 (tan (x) - 4 sec (x)) + C

Solution steps

\int 8 sec (x) (sec (x) - 4 tan (x)) d x

Take the constant out:

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

= 8 \cdot \int sec (x) (sec (x) - 4 tan (x)) d x

Rewrite using trig identities

= 8 \cdot \int \frac{1 - 4 sin ( x )}{cos ^{2} ( x )} d x

Expand

\frac{1 - 4 sin ( x )}{cos ^{2} ( x )} : \frac{1}{cos ^{2} ( x )} - \frac{4 sin ( x )}{cos ^{2} ( x )}

= 8 \cdot \int \frac{1}{cos ^{2} ( x )} - \frac{4 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

= 8 (\int \frac{1}{cos ^{2} ( x )} d x - \int \frac{4 sin ( x )}{cos ^{2} ( x )} d x)

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

\int \frac{4 sin ( x )}{cos ^{2} ( x )} d x = 4 sec (x)

= 8 (tan (x) - 4 sec (x))

Add a constant to the solution

= 8 (tan (x) - 4 sec (x)) + C

y^{^{''}} + 8 y^{^{'}} + 4 y = t

\frac{\partial}{\partial y} (sin (z))

4 x + 9, 144 - x, - 9, 144

\frac{2}{x ^{3}} - \frac{8}{x}

y = (\frac{f ( x )}{g ( x )})^{2}

What is the integral of 8sec(x)(sec(x)-4tan(x)) ?
The integral of 8sec(x)(sec(x)-4tan(x)) is 8(tan(x)-4sec(x))+C