Question

$\int \sec^n x \tan x dx$ is equal to

• A. $\frac{\sec^n x}{n} + c$
• B. $\frac{\sec^2 x}{2} + c$
• C. $\frac{\tan x}{n} + c$
• D. $\frac{(\sec^n x)}{n}$

Answer 1

Answer: (A)

(a) $\frac{\sec^n x}{n} + c$

Answer 2

To evaluate the integral $\int \sec^n x \tan x dx$, we can use the method of substitution.

Let $u = \sec x$.
Now, differentiate $u$ with respect to $x$:
$\frac{du}{dx} = \frac{d}{dx}(\sec x) = \sec x \tan x$.
From this, we can write $du = \sec x \tan x dx$.

Now, rewrite the original integral by separating one $\sec x$ term:
$\int \sec^n x \tan x dx = \int \sec^{n-1} x \cdot (\sec x \tan x) dx$.

Substitute $u = \sec x$ and $du = \sec x \tan x dx$ into the integral:
The integral becomes $\int u^{n-1} du$.

Now, apply the power rule for integration, which states that $\int x^k dx = \frac{x^{k+1}}{k+1} + C$ (for $k \neq -1$).
In our case, $k = n-1$. So,
$\int u^{n-1} du = \frac{u^{(n-1)+1}}{(n-1)+1} + C = \frac{u^n}{n} + C$.

Finally, substitute back $u = \sec x$:
The integral evaluates to $\frac{(\sec x)^n}{n} + C = \frac{\sec^n x}{n} + C$.

This result is valid for $n \neq 0$.

Comparing this result with the given options:
(a) $\frac{\sec^n x}{n} + c$
(b) $\frac{\sec^2 x}{2} + c$
(c) $\frac{\tan x}{n} + c$
(d) $\frac{(\sec^n x)}{n}$

Option (a) matches our derived solution exactly, including the constant of integration 'c'. Option (d) is missing the constant of integration, which is conventionally included in indefinite integrals.