Calculus Help!

[dux]

how to integrate this:

f(x) = sec(x)[ 8tan(x) - 5sec(x) ]

 

?

[kyulkyung]

o god i forgot all my calc 

[ricecake]

multiply it out 

 

f(x)=8secxtanx - 5sec2x 

pull out the constants

integrate (secxtanx --> secx; sec2x --> tanx) 

 

edit: so the answer should be 8secx - 5tanx + c i believe

[Stepkdramafan226]

https://www.wolframalpha.com/input/?i=f(x)+%3D+sec(x)%5B+8tan(x)+-+5sec(x)+%5D

 

https://www.wolframalpha.com/input/?i=integrate+f(x)+%3D+sec(x)%5B+8tan(x)+-+5sec(x)+%5D

 

Forgot about calculus when I didn't need it anymore. 

Not sure if that helps tbh 

[Swirl900]

Take the integral:
integral sec(x) (8 tan(x) - 5 sec(x)) dx

Expanding the integrand sec(x) (8 tan(x) - 5 sec(x)) gives 8 tan(x) sec(x) - 5 sec^2(x):
= integral(8 tan(x) sec(x) - 5 sec^2(x)) dx

Integrate the sum term by term and factor out constants:
= 8 integral tan(x) sec(x) dx - 5 integral sec^2(x) dx

Rewrite tan(x) sec(x) as (sin(x))/(cos^2(x)):
= 8 integral(sin(x))/(cos^2(x)) dx - 5 integral sec^2(x) dx

For the integrand (sin(x))/(cos^2(x)), substitute u = cos(x) and du = -sin(x) dx:
= 8 integral-1/u^2 du - 5 integral sec^2(x) dx

Factor out constants:
= -8 integral1/u^2 du - 5 integral sec^2(x) dx

The integral of 1/u^2 is -1/u:
= 8/u - 5 integral sec^2(x) dx

The integral of sec^2(x) is tan(x):
= 8/u - 5 tan(x) + constant

Substitute back for u = cos(x):
Answer: |
| = 8 sec(x) - 5 tan(x) + constant

 

 

 

I copied that from wolfram ^^^

[Izara]

Problem:

∫sec(x)(8tan(x)−5sec(x))dx∫secâ¡(x)(8tanâ¡(x)−5secâ¡(x))dx

Expand:

=∫(8sec(x)tan(x)−5sec2(x))dx=∫(8secâ¡(x)tanâ¡(x)−5sec2â¡(x))dx

Apply linearity:

=8∫sec(x)tan(x)dx−5∫sec2(x)dx=8∫secâ¡(x)tanâ¡(x)dx−5∫sec2â¡(x)dx

Now solving:

∫sec(x)tan(x)dx∫secâ¡(x)tanâ¡(x)dx

This is a standard integral:

=sec(x)=secâ¡(x)

Now solving:

∫sec2(x)dx∫sec2â¡(x)dx

This is a standard integral:

=tan(x)=tanâ¡(x)

Plug in solved integrals:

8∫sec(x)tan(x)dx−5∫sec2(x)dx8∫secâ¡(x)tanâ¡(x)dx−5∫sec2â¡(x)dx

=8sec(x)−5tan(x)=8secâ¡(x)−5tanâ¡(x)

The problem is solved:

∫sec(x)(8tan(x)−5sec(x))dx∫secâ¡(x)(8tanâ¡(x)−5secâ¡(x))dx

=8sec(x)−5tan(x)+C=8secâ¡(x)−5tanâ¡(x)+C

ANTIDERIVATIVE COMPUTED BY MAXIMA:

∫f(x)dx=F(x)=∫f(x)dx=F(x)=

8cos(x)−5tan(x)+C8cosâ¡(x)−5tanâ¡(x)+C

No further simplification found!

[Pleasure]

f(x) = pink bible

 

there u go boo <3

[dux]

multiply it out 

 

f(x)=8secxtanx - 5sec2x 

pull out the constants

integrate (secxtanx --> secx; sec2x --> tanx) 

 

edit: so the answer should be 8secx - 5tanx + c i believe

 

 

Take the integral:

integral sec(x) (8 tan(x) - 5 sec(x)) dx

 

Expanding the integrand sec(x) (8 tan(x) - 5 sec(x)) gives 8 tan(x) sec(x) - 5 sec^2(x):

= integral(8 tan(x) sec(x) - 5 sec^2(x)) dx

 

Integrate the sum term by term and factor out constants:

= 8 integral tan(x) sec(x) dx - 5 integral sec^2(x) dx

 

Rewrite tan(x) sec(x) as (sin(x))/(cos^2(x)):

= 8 integral(sin(x))/(cos^2(x)) dx - 5 integral sec^2(x) dx

 

For the integrand (sin(x))/(cos^2(x)), substitute u = cos(x) and du = -sin(x) dx:

= 8 integral-1/u^2 du - 5 integral sec^2(x) dx

 

Factor out constants:

= -8 integral1/u^2 du - 5 integral sec^2(x) dx

 

The integral of 1/u^2 is -1/u:

= 8/u - 5 integral sec^2(x) dx

 

The integral of sec^2(x) is tan(x):

= 8/u - 5 tan(x) + constant

 

Substitute back for u = cos(x):

Answer: |

| = 8 sec(x) - 5 tan(x) + constant

 

 

 

I copied that from wolfram ^^^

 

 

 

Problem:

∫sec(x)(8tan(x)−5sec(x))dx∫secâ¡(x)(8tanâ¡(x)−5secâ¡(x))dx

Expand:

=∫(8sec(x)tan(x)−5sec2(x))dx=∫(8secâ¡(x)tanâ¡(x)−5sec2â¡(x))dx

Apply linearity:

=8∫sec(x)tan(x)dx−5∫sec2(x)dx=8∫secâ¡(x)tanâ¡(x)dx−5∫sec2â¡(x)dx

Now solving:

∫sec(x)tan(x)dx∫secâ¡(x)tanâ¡(x)dx

This is a standard integral:

=sec(x)=secâ¡(x)

Now solving:

∫sec2(x)dx∫sec2â¡(x)dx

This is a standard integral:

=tan(x)=tanâ¡(x)

Plug in solved integrals:

8∫sec(x)tan(x)dx−5∫sec2(x)dx8∫secâ¡(x)tanâ¡(x)dx−5∫sec2â¡(x)dx

=8sec(x)−5tan(x)=8secâ¡(x)−5tanâ¡(x)

The problem is solved:

∫sec(x)(8tan(x)−5sec(x))dx∫secâ¡(x)(8tanâ¡(x)−5secâ¡(x))dx

=8sec(x)−5tan(x)+C=8secâ¡(x)−5tanâ¡(x)+C

ANTIDERIVATIVE COMPUTED BY MAXIMA:

∫f(x)dx=F(x)=∫f(x)dx=F(x)=

8cos(x)−5tan(x)+C8cosâ¡(x)−5tanâ¡(x)+C

 

 

No further simplification found!

 

thank you all so much! i idndt know wolfram did that!

[Izara]

thank you all so much! i idndt know wolfram did that!

Here you go..

https://www.integral-calculator.com

just put yr question that you wanna integrate and the calculator will show you step by step

 

I had done my study in Calculus 3 years ago..so I can't really remembered it