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

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

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integral of tan(x)(4cos(x)+8sec(x))

\int tan (x) (4 cos (x) + 8 sec (x)) d x

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

Solution steps

\int tan (x) (4 cos (x) + 8 sec (x)) d x

Rewrite using trig identities

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

Expand

\frac{sin ( x ) ( 4 cos ^{2} ( x ) + 8 )}{cos ^{2} ( x )} : 4 sin (x) + \frac{8 sin ( x )}{cos ^{2} ( x )}

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

\int 4 sin (x) d x = - 4 cos (x)

\int \frac{8 sin ( x )}{cos ^{2} ( x )} d x = 8 sec (x)

= - 4 cos (x) + 8 sec (x)

Add a constant to the solution

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

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d er i v a t i v e f (t) = 5 - t

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