How do you solve this calculus 2 integration problem?

[harpazo]

I am currently studying calculus 3. However, once in a while, I like to review single variable calculus. I decided to tackle the following calculus 2 integration problem:

∫ tan^3 x dx

Solution:

Separate into two tangent functions.

(tan^2 x)(tan x)

Rewrite tan^2 x as sec^2 x - 1 using its equivalent trig identity.

∫ (tan x)(sec^2 x - 1) dx

Let u = sec^2 x

du/dx = tan x

dx = du/ tan x

∫ (tan x)(u^2 - 1) du/tan x

∫ (u^2 - 1) du

(u^3/3) - u + C

Back-substitute for u.

We know that u = sec^2 x.

[(sec^2 x)^3]/3 - sec^2 x + C

What do you say?

 

[MarkFL]

If you differentiate your result w.r.t $x$, do you get your original integrand as the result?

 

[harpazo]

If you differentiate your result w.r.t $x$, do you get your original integrand as the result?

No. I did not get back the original problem. Can you go through my reply to see where I went wrong?

 

[MarkFL]

No. I did not get back the original problem. Can you go through my reply to see where I went wrong?

We are given:

$$I=\int\tan^3(x)\,dx$$

I agree that a good strategy is to write this as:

$$I=\int\tan^2(x)\cdot\tan(x)\,dx$$

Now we can use the Pythagorean identity:

$$\tan^2(\theta)=\sec^2(\theta)-1$$

To write the integral as:

$$I=\int\tan(x)\left(\sec^2(x)-1\right)\,dx$$

Distribute:

$$I=\int\tan(x)\sec^2(x)-\tan(x)\,dx=\int \tan(x)\,\sec^2(x)\,dx+\int \frac{-\sin(x)}{\cos(x)}\,dx$$

Can you now spot the appropriate substitutions to be used on the two integrals?

 

[harpazo]

∫ (tan x)(sec^2 x) dx + ∫ -(sin x)/(cos x) dx

For the left side integral, let u = tan x

du/dx = sec^2 x

dx = du/sec^2 x

∫ u• sec^2 x • du/sec^2 x

∫ u du + ∫ -(sin x)/(cos x) dx

For the integral on the right side,
let u = cos x

du/dx = - sin x

dx = du/-sin x

∫ u du + ∫ -(sin x)/u • du/-sin x

∫ u du + ∫ 1/u • du

(u^2)/2 + ln u + C

Back-substitute for u.

[(sec^2 x)^2]/2 + ln (cos x) + C

 

[MarkFL]

∫ (tan x)(sec^2 x) dx + ∫ -(sin x)/(cos x) dx

For the left side integral, let u = tan x

du/dx = sec^2 x

dx = du/sec^2 x

∫ u• sec^2 x • du/sec^2 x

∫ u du + ∫ -(sin x)/(cos x) dx

For the integral on the right side,
let u = cos x

du/dx = - sin x

dx = du/-sin x

∫ u du + ∫ -(sin x)/u • du/-sin x

∫ u du + ∫ 1/u • du

(u^2)/2 + ln u + C

Back-substitute for u.

[(sec^2 x)^2]/2 + ln (cos x) + C

For the first integral, you let:

$$u=\tan(x)$$

And so you would have:

$$I=\frac{1}{2}\tan^2(x)+\ln|\cos(x)|+C$$

Now, suppose that for the first integral, we let:

$$u=\sec(x)\implies du=\sec(x)\tan(x)\,dx$$

And so we eventually get:

$$I=\frac{1}{2}\sec^2(x)+\ln|\cos(x)|+C$$

How do we justify that both are correct? (Thinking)

 

[harpazo]

For the first integral, you let:

$$u=\tan(x)$$

And so you would have:

$$I=\frac{1}{2}\tan^2(x)+\ln|\cos(x)|+C$$

Now, suppose that for the first integral, we let:

$$u=\sec(x)\implies du=\sec(x)\tan(x)\,dx$$

And so we eventually get:

$$I=\frac{1}{2}\sec^2(x)+\ln|\cos(x)|+C$$

How do we justify that both are correct? (Thinking)

We take the derivative to check our answer.

 

[MarkFL]

We take the derivative to check our answer.

Well, yes differentiating both forms gets us back to the original integrand, but how do we "justify" that both are correct?

 

[harpazo]

Well, yes differentiating both forms gets us back to the original integrand, but how do we "justify" that both are correct?

I do not understand your question.

 

[MarkFL]

I do not understand your question.

In post #6, I gave two different anti-derivatives...how can we show both are correct?

 

[harpazo]

In post #6, I gave two different anti-derivatives...how can we show both are correct?

We can show they are both correct by setting them equal to each other.

 

[JorgeLuna]

For the first integral, you let:

$$u=\tan(x)$$

And so you would have:

$$I=\frac{1}{2}\tan^2(x)+\ln|\cos(x)|+C$$

Now, suppose that for the first integral, we let:

$$u=\sec(x)\implies du=\sec(x)\tan(x)\,dx$$

And so we eventually get:

$$I=\frac{1}{2}\sec^2(x)+\ln|\cos(x)|+C$$

How do we justify that both are correct? (Thinking)

Would showing that the two anti-derivative forms differ by a constant suffice as justification?

$$I=\frac{1}{2}\tan^2(x)+\ln|\cos(x)|+C$$

$$I=\frac{1}{2}[\sec^2(x)-1]+\ln|\cos(x)|+C$$

$$I=\frac{1}{2}\sec^2(x)-\frac{1}{2}+\ln|\cos(x)|+C$$

$$I=\frac{1}{2}\sec^2(x)+\ln|\cos(x)|+C_2$$

 

[MarkFL]

Would showing that the two anti-derivative forms differ by a constant suffice as justification?

$$I=\frac{1}{2}\tan^2(x)+\ln|\cos(x)|+C$$

$$I=\frac{1}{2}[\sec^2(x)-1]+\ln|\cos(x)|+C$$

$$I=\frac{1}{2}\sec^2(x)-\frac{1}{2}+\ln|\cos(x)|+C$$

$$I=\frac{1}{2}\sec^2(x)+\ln|\cos(x)|+C_2$$

Yes, that was exactly the kind of thing I was hoping the OP would state. (Yes)

 

[harpazo]

I don't know...