Question

Evaluate the integral. (Use C for the constant of integration.)

Answer 1

\[\int\limits 4 \tan^3x \sec x dx\]

Answer 2

Here is what i did.

Answer 3

\[4 \int\limits \tan x \sec x (\tan^2 x) dx\]
\[4 \int\limits \tan x \sec x (\sec^2x -1) dx\]
u = sec x du=secxtanx
\[4 \int\limits (u^2 -1) du\]
\[4 \left[ \frac{ u^3 }{ 3 } -x \right]\]
\[\frac{ 4 \sec^3x }{ 3 } - 4x\]

Answer 4

So my answer was \[\frac{ 4 \sec^3 x }{ 3 } - 4x +C\]

But it is saying that my answer is wrong. Can someone please check my work and answer. I only have 1 more attempt at the right answer.

Answer 5

line 5

you put x in instead of u

Answer 6

\[4\left[ \int\limits u^2 - \int\limits 1\right]\]

Answer 7

Dont you do that?

Answer 8

the integral of 1 du = u

Answer 9

Oh okay.

Answer 10

yw

Answer 11

Thanks. lol

Answer 12

no probs