Q. \[ \int \tan^2 x \, dx \]

Q: How do you simplify \tan^2 x before integrating?

Use \tan^2 x=\sec^2 x-1 . Then integrate \int \tan^2 x\,dx=\int(\sec^2 x-1)\,dx=\tan x-x+C .

Q: What is \int \sec^2 x\,dx ?

Since \frac{d}{dx}(\tan x)=\sec^2 x , we have \int \sec^2 x\,dx=\tan x+C .

Q: How do you integrate \tan^2 x using a trig identity to avoid substitution?

Convert \tan^2 x=\sec^2 x-1 . Then \int(\sec^2 x-1)\,dx=\tan x-x+C .

Q: What derivative confirms the result \frac{d}{dx}(\tan x-x)=\tan^2 x ?

Compute \frac{d}{dx}(\tan x-x)=\sec^2 x-1=\tan^2 x .

Q: Is there an alternative result using \tan^2 x=\frac{1}{\cos^2 x}-1 ?

Yes. \tan^2 x=\frac{1}{\cos^2 x}-1=\sec^2 x-1 . Thus \int \tan^2 x\,dx=\tan x-x+C .

Answer

To integrate \( \tan^2 x \), use the identity \( \tan^2 x = \sec^2 x – 1 \).

\[
\int \tan^2 x\,dx=\int (\sec^2 x-1)\,dx=\int \sec^2 x\,dx-\int 1\,dx
\]
\[
=\tan x – x + C
\]

Final result: \(\tan x – x + C\).

Detailed Explanation

We want to compute the indefinite integral

\[
\int \tan^2 x \, dx
\]

Step 1: Rewrite \(\tan^2 x\) using a trigonometric identity.

Use the identity

\[
1+\tan^2 x = \sec^2 x
\]

Rearrange it to express \(\tan^2 x\) in terms of \(\sec^2 x\):

\[
\tan^2 x = \sec^2 x – 1
\]

Step 2: Substitute into the integral.

Replace \(\tan^2 x\) with \(\sec^2 x – 1\):

\[
\int \tan^2 x \, dx = \int (\sec^2 x – 1)\, dx
\]

Step 3: Split the integral into two simpler integrals.

Use linearity of integration:

\[
\int (\sec^2 x – 1)\, dx = \int \sec^2 x \, dx – \int 1 \, dx
\]

Step 4: Integrate each part.

First, use the standard result:

\[
\int \sec^2 x \, dx = \tan x
\]

Second, integrate \(1\):

\[
\int 1\, dx = x
\]

Step 5: Combine results and add the constant of integration.

So the integral becomes:

\[
\int \tan^2 x \, dx = \tan x – x + C
\]

Final Answer:

\[
\boxed{\int \tan^2 x \, dx = \tan x – x + C}
\]

See full solution

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Calculus FAQ

How do you simplify \( \tan^2 x \) before integrating?

Use \( \tan^2 x=\sec^2 x-1 \). Then integrate \( \int \tan^2 x\,dx=\int(\sec^2 x-1)\,dx=\tan x-x+C \).

What is \( \int \sec^2 x\,dx \)?

Since \( \frac{d}{dx}(\tan x)=\sec^2 x \), we have \( \int \sec^2 x\,dx=\tan x+C \).

How do you integrate \( \tan^2 x \) using a trig identity to avoid substitution?

Convert \( \tan^2 x=\sec^2 x-1 \). Then \( \int(\sec^2 x-1)\,dx=\tan x-x+C \).

Can you integrate \( \tan^2 x \) by rewriting in terms of \( \sin x \) and \( \cos x \)?

\( \tan^2 x=\frac{\sin^2 x}{\cos^2 x}=\frac{1-\cos^2 x}{\cos^2 x}=\sec^2 x-1 \). So \( \int \tan^2 x\,dx=\tan x-x+C \).

What derivative confirms the result \( \frac{d}{dx}(\tan x-x)=\tan^2 x \)?

Compute \( \frac{d}{dx}(\tan x-x)=\sec^2 x-1=\tan^2 x \).

Is there an alternative result using \( \tan^2 x=\frac{1}{\cos^2 x}-1 \)?

Yes. \( \tan^2 x=\frac{1}{\cos^2 x}-1=\sec^2 x-1 \). Thus \( \int \tan^2 x\,dx=\tan x-x+C \).

What is the final integral \( \int \tan^2 x\,dx \)?

\( \int \tan^2 x\,dx=\tan x-x+C \).

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