Question

$\int \frac{sec^8x}{cosx}dx =$

• A. $\frac{sec^9x}{8}+C$
• B. $\frac{sec^7x}{7}+C$
• C. $\frac{sec^6x}{6}+C$
• D. $\frac{sec^9x}{9}+C$

Answer 1

Answer: (D)

$\frac{sec^9x}{9}+C$

Answer 2

The given integral is $\int \frac{sec^8x}{cosx}dx$.

First, simplify the integrand: We know that $\frac{1}{cosx} = secx$. So, the integrand becomes $sec^8x \cdot secx = sec^{8+1}x = sec^9x$. The integral is therefore $\int sec^9x dx$.

This integral is not a direct application of the power rule for integration, as $\int sec^n x dx$ typically involves reduction formulas or substitution with $\tanx$. For example, $\frac{d}{dx}(sec^nx) = n \cdot sec^{n-1}x \cdot (secx \tanx) = n \cdot sec^nx \tanx$.

Let's consider the possibility that there is a missing $\tanx$ term in the numerator of the integrand. If the integral was intended to be $\int \frac{sec^8x}{cosx} \tanx dx$, which simplifies to $\int sec^9x \tanx dx$. Let $u = secx$. Then $du = secx \tanx dx$. The integral can be rewritten as $\int sec^8x (secx \tanx) dx$. Substituting $u$ and $du$: $\int u^8 du = \frac{u^{8+1}}{8+1} + C = \frac{u^9}{9} + C$. Substitute back $u = secx$: $\frac{sec^9x}{9} + C$.

This result matches option (D). Given the simplicity of the options and the common structure of such problems in competitive exams where a slight typo might occur, it is highly probable that the question intended to include a $\tanx$ term in the numerator, making the integral solvable by a simple substitution. Without the $\tanx$ term, the integral $\int sec^9x dx$ is significantly more complex and does not yield any of the given options directly.