Question

evaluate the indefinite integral.
\\(\int \tan^{7}(x) \sec^{4}(x) \\, dx = \square + c\\).

Explanation:

Step1: Rewrite \(\sec^4(x)\)

We know that \(\sec^4(x)=\sec^2(x)\cdot\sec^2(x)\), and by the Pythagorean identity \(\sec^2(x) = 1+\tan^2(x)\). So we can rewrite the integral as:

$$ \int\tan^7(x)\sec^2(x)\cdot\sec^2(x)dx=\int\tan^7(x)(1 + \tan^2(x))\sec^2(x)dx $$

Step2: Substitution

Let \(u=\tan(x)\), then \(du=\sec^2(x)dx\). Substituting into the integral, we get:

$$ \int u^7(1 + u^2)du $$

Step3: Expand the integrand

Expand \(u^7(1 + u^2)\) to get \(u^7+u^9\). So the integral becomes:

$$ \int(u^7 + u^9)du $$

Step4: Integrate term by term

Using the power rule for integration \(\int x^n dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)), we have:

$$ \int u^7du+\int u^9du=\frac{u^8}{8}+\frac{u^{10}}{10}+C $$

Step5: Substitute back \(u = \tan(x)\)

Substitute \(u=\tan(x)\) back into the expression:

$$ \frac{\tan^8(x)}{8}+\frac{\tan^{10}(x)}{10}+C $$

Answer:

\(\frac{\tan^{10}(x)}{10}+\frac{\tan^8(x)}{8}\)