Evaluate the integral. (Remember the constant of integration.) $\int 7 \tan^3(x) \sec(x) \, dx$

Neetesh Kumar | December 7, 2024

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This is the solution to Math 132
Assignment: 7.2 Question Number 3
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Step-by-step solution:

We are tasked with solving the integral: $\int 7 \tan^3(x) \sec(x) \, dx$

Step 1: Simplify the powers of $\tan(x)$

We rewrite $\tan^3(x)$ as $\tan^2(x) \tan(x)$:

$\int 7 \tan^3(x) \sec(x) \, dx = \int 7 \tan^2(x) \tan(x) \sec(x) \, dx$

Using the trigonometric identity $\tan^2(x) = \sec^2(x) - 1$, substitute:

$\int 7 \tan^3(x) \sec(x) \, dx = \int 7 (\sec^2(x) - 1) \tan(x) \sec(x) \, dx$

Step 2: Substitution

Let $u = \sec(x)$. Then:

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

Substitute $u = \sec(x)$ and $du = \sec(x) \tan(x) \, dx$ into the integral:

$\int 7 (\sec^2(x) - 1) \tan(x) \sec(x) \, dx = \int 7 (u^2 - 1) \, du$

Step 3: Solve the integral

Expand and solve the integral:

$\int 7 (u^2 - 1) \, du = 7 \int u^2 \, du - 7 \int 1 \, du$

1. Solve $7 \int u^2 \, du$: $\int u^2 \, du = \frac{u^3}{3} \implies 7 \int u^2 \, du = \frac{7 u^3}{3}$

2. Solve $-7 \int 1 \, du$: $\int 1 \, du = u \implies -7 \int 1 \, du = -7 u$

Combine the results: $\int 7 (u^2 - 1) \, du = \frac{7 u^3}{3} - 7 u + C$

Step 4: Substitute back $u = \sec(x)$

Replace $u$ with $\sec(x)$:

$\frac{7 u^3}{3} - 7 u + C = \frac{7 \sec^3(x)}{3} - 7 \sec(x) + C$

Final Answer

$\int 7 \tan^3(x) \sec(x) \, dx = \boxed{\frac{7 \sec^3(x)}{3} - 7 \sec(x) + C}$

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