Question

Reduction formulas can be used to compute integrals of higher power of sin x, cos x, tan x etc. $\int sec^6 x dx = \frac{1}{5}tan^5 x + A tan^3 x + tan x + C$, then:

• A. $A = \frac{1}{3}$
• B. $A = \frac{2}{3}$
• C. $A = -\frac{1}{3}$
• D. $A = -\frac{2}{3}$

Answer 1

Answer: (B)

$A = \frac{2}{3}$

Answer 2

1. Rewrite the integral: $\int sec^6 x dx = \int sec^4 x \cdot sec^2 x dx$.
2. Use the identity $sec^2 x = 1 + tan^2 x$: $\int (1 + tan^2 x)^2 \cdot sec^2 x dx$.
3. Substitute $u = tan x$, so $du = sec^2 x dx$. The integral becomes $\int (1 + u^2)^2 du$.
4. Expand the integrand: $\int (1 + 2u^2 + u^4) du$.
5. Integrate term by term: $u + \frac{2u^3}{3} + \frac{u^5}{5} + C$.
6. Substitute back $u = tan x$: $tan x + \frac{2}{3} tan^3 x + \frac{1}{5} tan^5 x + C$.
7. Compare this result with the given expression $\frac{1}{5}tan^5 x + A tan^3 x + tan x + C$.
8. Equating the coefficients of $tan^3 x$, we get $A = \frac{2}{3}$.