# More substitution

Gold Member

## Homework Statement

$$\int {\sec ^3 x\,\,\tan x\,\,dx}$$

## Homework Equations

$$u = \sec x$$
This is my guess at u.

## The Attempt at a Solution

$$\frac{{du}}{{dx}} = \sec x\,\,\tan x,\,\,\,dx = \frac{{du}}{{\sec x\,\,\tan x}}$$

$$\int {\sec ^3 x\,\,\tan x\,\,dx} = \int {u^3 \,\,\tan x\,\,\frac{{du}}{{\sec x\,\,\tan x}}} = \int {u^3 \,\,\tan x\,\,\frac{{du}}{{u\,\,\tan x}} = }$$

$$\int {u^3 \,\,\,\,\frac{{du}}{u}} = \int {\frac{{u^3 }}{u}} du = \int {u^2 du} = 2u + C =$$

$$2\sec x + C$$

But the back of the book says
$$\frac{1}{3}\sec ^3 x + C$$

Playing around with it on the graphing software, I find that my answer and the book's answer produce different graphs, however, they both seem to have the problem stated in the question as their derivatives. Is one answer more correct than the other?

$$\frac{1}{3}\sec ^3 x + C$$ $$2\sec x + C$$ ## The Attempt at a Solution

AKG
Homework Helper
The integral of u2 is u3/3, not 2u. You differentiated instead of integrating.

Gold Member
oops. Thanks for catching that.

Gib Z
Homework Helper
If the derivatives are equal, the functions can only differ by a constant. Do not accept answers as 'more correct', its right or wrong.

Edit: I just checked the post again, your working is alot more than I'd have thought >.< Remember how substitution should be used when the function and its derivative is there? Rewrite the integral as:

$$\int \sec^2 x \sec x \tan x dx$$ Then Letting u= sec x, $$\int u^2 du$$, which seems faster to me ...just a tip.

Last edited: