Let I_n = int sec^n x , dx. If 5 I_6 - 4 I_4 | vedclass

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(B) Given that $I_n = \int \sec^n x \, dx$. $f(x) = 5 I_6 - 4 I_4 = 5 \int \sec^6 x \, dx - 4 \int \sec^4 x \, dx$. $f(x) = \int (5 \sec^4 x \cdot \se

Vedclass · Accepted Answer

$4$

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(B) Given that $I_n = \int \sec^n x \, dx$. $f(x) = 5 I_6 - 4 I_4 = 5 \int \sec^6 x \, dx - 4 \int \sec^4 x \, dx$. $f(x) = \int (5 \sec^4 x \cdot \sec^2 x - 4 \sec^2 x \cdot \sec^2 x) \, dx$. Using $\sec^2 x = 1 + 	an^2 x$, we get: $f(x) = \int \{5(1 + 	an^2 x)^2 - 4(1 + 	an^2 x)\} \sec^2 x \, dx$. Let $u = 	an x$, then $du = \sec^2 x \, dx$. $f(x) = \int \{5(1 + u^2)^2 - 4(1 + u^2)\} \, du$. $f(x) = \int \{5(1 + u^4 + 2u^2) - 4 - 4u^2\} \, du$. $f(x) = \int (5 + 5u^4 + 10u^2 - 4 - 4u^2) \, du = \int (5u^4 + 6u^2 + 1) \, du$. $f(x) = u^5 + 2u^3 + u = 	an^5 x + 2 	an^3 x + 	an x$. At $x = \frac{\pi}{4}$, $	an\left(\frac{\pi}{4}ight) = 1$. $f\left(\frac{\pi}{4}ight) = (1)^5 + 2(1)^3 + 1 = 1 + 2 + 1 = 4$.

Let $I_n = \int \sec^n x \, dx$. If $5 I_6 - 4 I_4 = f(x)$, then $f\left(\frac{\pi}{4}\right)$ is equal to

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