Where is my mistake in this problem (finding indefinite integral using trig identities)

Question

I am learning calculus. Currently I am studying integration using trigonometric identities.

I encountered this problem and gave it a try. My answer was incorrect. I then sought the correct answer, and I see now why it is correct and why the method used to solve it was better than mine. Here is my issue: I can not for the life of me find where my error was. Why is this answer wrong? I numbered the steps and explained my thinking. Here is the problem:

$$\int \tan^5(x)\cdot \sec^4(x) \,dx$$

I began by looking over my list of trig identities and derivatives, and I found these two which appeared helpful. My thinking was that I could get the derivative of $\tan(x)$ and then use the identity to express everything that remains in terms of secants to use u-substitution.

$\sec^2(x)-1=\tan^2(x)$ and $\frac{d}{dx}\sec(x)=\tan(x)\cdot sec(x)$

First I got $\tan x \cdot \sec x$ by itself:

1. $$\int \left[\tan^5(x)\cdot \sec^4(x) \right] \,dx = \int \left[\tan^4(x)\cdot \sec^3(x)\cdot \tan(x)\cdot \sec(x) \right] dx$$

Then I broke the remaining tangent terms into $\tan^2x$ terms:

2. $$= \int \left[\tan^2(x)\cdot \tan^2(x) \cdot \sec^3(x)\cdot \tan(x)\cdot \sec(x) \right] \,dx$$

Then I replaced each $\tan^2x$ term by its equivalent term in secants:

3. $$=\int \left[ \left(\sec^2(x)-1 \right)\cdot \left(\sec^2(x)-1 \right)\cdot \sec^3(x)\cdot \tan(x)\cdot \sec(x) \right] \,dx$$

Then I distributed all my secant terms:

4. $$=\int \left[(\sec^4(x)-2 \sec^2(x)+1)(\sec^3(x))\cdot \tan(x)\cdot \sec(x) \right] \,dx$$
5. $$=\int \left[ \left(\sec^7(x)-2 \sec^5(x)+ \sec^3(x) \right)\cdot \tan(x)\cdot \sec(x) \right] dx$$

Then I applied this substitution:

6. $u=\sec(x)$, and then $\,du=\tan(x)\sec(x) dx$

Applying the substitution:

7. $$=\int \left(u^7-2u^5+u^3 \right) \,du$$

Integrating:

8. $$=\frac{u^8}{8}-\frac{2u^6}{6}+\frac{u^4}{4}+C$$

And then substituting back in:

9. $$=\frac{1}{8} \sec^8(x)-\frac{1}{3} \sec^6(x)+\frac{1}{4} \sec^4(x)+C$$

Of course, this answer is wrong. I would love to know where I went wrong!

EDIT: I forgot to say, please forgive errors in the mathjax -- I'm a rank beginner!

Answer 1

$$\int \tan^5 \theta \cdot \sec^4\theta\, d\theta = \int \tan^5\theta(1+\tan^2 \theta)\sec^2\theta \, d\theta$$

Doing the substitution $u=\tan \theta$ ($du=\sec^2\theta\, d\theta$), we get:

$$\int u^5(1+u^2)du=\dfrac{u^6}{6}+\dfrac{u^8}{8}+C=\dfrac{\tan^6\theta}{6}+\dfrac{\tan^8 \theta}{8}+C$$

Answer 2

Who says you went wrong? Subtract the two answers

$$\frac{1}{8}\sec^8x-\frac{1}{8}\tan^8x-\frac{1}{3}\sec^6x-\frac{1}{6}\tan^6x+\frac{1}{4}\sec^4x$$

$$= \frac{(\sec^2x+\tan^2x)(\sec^4x+\tan^4x)}{8} - \frac{\sec^6x}{2} + \frac{(\sec^4x+\sec^2x\tan^2x+\tan^4x)}{6} + \frac{\sec^4x}{4}$$

which is $\frac{1}{24}$, a constant.

Answer 3

You were told the correct answer is $$ \int \tan^5 x \sec^4 x \,\mathrm dx = \frac{\tan^8 x}{8} + \frac{\tan^6 x}{6} + C. $$

You know that $\tan^2 x = \sec^2 x - 1$. Therefore $\tan^6 x = (\sec^2 x - 1)^3$ and $\tan^8 x = (\sec^2 x - 1)^4$. Therefore

$$ \frac{\tan^8 x}{8} + \frac{\tan^6 x}{6} = \frac{(\sec^2 x - 1)^4}{8} + \frac{(\sec^2 x - 1)^3}{6}. $$

Expand the powers of $\sec^2 x - 1$ and combine like terms. You will find that

$$ \frac{(\sec^2 x - 1)^4}{8} + \frac{(\sec^2 x - 1)^3}{6} = \frac{\sec^8 x}{8} - \frac{\sec^6 x}{3} + \frac{\sec^4 x}{4} - \frac{1}{24}, $$

so your answer is correct; all that is different is the constant of integration.