Question

$\int \sec^4 x \cdot \tan^4 x dx = \frac{\tan^m x}{m} + \frac{\tan^n x}{n} + c$ (where $c$ is constant of integration), then $m + n =$

• A. 8
• B. 12
• C. 10
• D. 16

Hint:

Whenever the exponent of $\sec x$ is a positive even integer, the standard algorithmic approach is always to peel off $\sec^2 x$ for $du$, change everything else to $\tan x$, and set $u = \tan x$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to evaluate a trigonometric integral, express it in a specific polynomial format of tangents, identify the powers $m$ and $n$, and calculate their sum.

Step 2: Key Formula or Approach:
For integrals involving powers of secant and tangent where the power of secant is even, we split off a $\sec^2 x$ term to serve as $dx$ for a $u$-substitution and convert the remaining secants to tangents using the identity:
$$\sec^2 x = 1 + \tan^2 x$$

Step 3: Detailed Explanation:
Let the given integral be $I$:
$$I = \int \sec^4 x \cdot \tan^4 x dx$$ Split $\sec^4 x$ into $\sec^2 x \cdot \sec^2 x$:
$$I = \int \sec^2 x \cdot \tan^4 x \cdot \sec^2 x dx$$ Apply the trigonometric identity to the first secant term:
$$I = \int (1 + \tan^2 x) \tan^4 x \cdot \sec^2 x dx$$ Now, perform the substitution $t = \tan x$. This gives $dt = \sec^2 x dx$:
$$I = \int (1 + t^2) t^4 dt$$ Distribute $t^4$ inside the parenthesis:
$$I = \int (t^4 + t^6) dt$$ Integrate term by term using the power rule:
$$I = \frac{t^5}{5} + \frac{t^7}{7} + c$$ Substitute $t = \tan x$ back into the expression:
$$I = \frac{\tan^5 x}{5} + \frac{\tan^7 x}{7} + c$$ Compare this result to the given format $\frac{\tan^m x}{m} + \frac{\tan^n x}{n} + c$. We can clearly see that:
$m = 5$ and $n = 7$ (the order does not matter).
The question asks for the sum $m + n$:
$$m + n = 5 + 7 = 12$$

Step 4: Final Answer:
The value of $m + n$ is 12, matching option (B).