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: Write the integral.
\[ \int \sec^4 x \tan^4 x\, dx \]
Step 2: Split one $\sec^2 x$ for substitution.
Keep one $\sec^2 x$ aside, since its derivative pairs with $\tan x$. Write $\sec^4 x = \sec^2 x \cdot \sec^2 x$ and use $\sec^2 x = 1 + \tan^2 x$.
\[ \int (1 + \tan^2 x)\tan^4 x \cdot \sec^2 x\, dx \]
Step 3: Substitute $t = \tan x$.
Then $dt = \sec^2 x\, dx$.
\[ \int (1 + t^2) t^4\, dt = \int (t^4 + t^6)\, dt \]
Step 4: Integrate term by term.
\[ = \frac{t^5}{5} + \frac{t^7}{7} + c \]
Step 5: Put $t = \tan x$ back.
\[ = \frac{\tan^5 x}{5} + \frac{\tan^7 x}{7} + c \]
Step 6: Read off $m$ and $n$.
Comparing with $\frac{\tan^m x}{m} + \frac{\tan^n x}{n}$, we get $m = 5$ and $n = 7$, so $m + n = 12$. \[ \boxed{m + n = 12 \text{ (Option 2)}} \]