Question

If $\int \tan^4 x dx = a \tan^3 x + b \tan x + cx + k$ (where k is the constant of integration) then the value of $a - b + c =$

• A. $\frac{7}{3}$
• B. $\frac{5}{3}$
• C. $\frac{4}{3}$
• D. $\frac{1}{3}$

Hint:

For $\int \tan^n x dx$, always pull out a $\tan^2 x$ and replace it with $(\sec^2 x - 1)$.

Answer 1

The Correct Option is B

Step 1: Concept
Use the trigonometric identity $\tan^2 x = \sec^2 x - 1$ to break down the integral.

Step 2: Meaning
$\int \tan^4 x dx = \int \tan^2 x (\sec^2 x - 1) dx = \int \tan^2 x \sec^2 x dx - \int \tan^2 x dx$.

Step 3: Analysis
1) $\int \tan^2 x \sec^2 x dx = \frac{\tan^3 x}{3}$.
2) $\int \tan^2 x dx = \int (\sec^2 x - 1) dx = \tan x - x$.
Total integral: $\frac{1}{3}\tan^3 x - \tan x + x + k$.
Comparing with given form: $a = 1/3, b = -1, c = 1$.

Step 4: Conclusion
$a - b + c = 1/3 - (-1) + 1 = 1/3 + 2 = 7/3$. Final Answer: (B)