Question

ASAP!!

Answer 1

✒️REDUCTION

Given:

$\maltese \: \boxed{ \boxed{ \displaystyle \tt \int \sec^n(u)\, du=\frac{\sec^{n-2}(u) \tan(u)}{n-1}+\frac{n-2}{n-1}\int \sec^{n-2}(u)\, du, \; n\neq1}}$

$\Large \mathcal{SOLUTION:}$

$\small \begin{array}{l} \normalsize \bold{Reduction\ Formula\ for}\: \displaystyle \tt \int \sec^n(u)\, du \\ \\ \displaystyle \tt \int \sec^n(u)\, du=\frac{\sec^{n-2}(u)\tan(u)}{n-1}+\frac{n-2}{n-1}\int \sec^{n-2}(u)\, du, \; n\neq1 \\ \\ \texttt{Verifying,} \\ \\ \texttt{Let } \displaystyle \tt I_n = \int \sec^n(u)\, du \\ \\ \displaystyle \tt I_n = \int \sec^{n-2}(u)\left(1 + \tan^2 (u)\right)\, du \\ \\ \displaystyle \tt I_n = \int \sec^{n-2}(u)\, du + \int \sec^{n - 2}(u) \tan^2 (u)\, du \\ \\ \displaystyle \tt I_n = \int \sec^{n-2}(u)\, du + \int \sec^{n - 2}(u) \tan^2 (u)\, du \\ \\ \displaystyle \tt I_n = \int \sec^{n-2}(u)\, du + \int \dfrac{1}{\cos^{n - 2}(u)} \cdot \dfrac{\sin^2 (u)}{\cos^2 (u)}\, du \\ \\ \displaystyle \tt I_n = \int \sec^{n-2}(u)\, du + \int \sin (u) \cdot \dfrac{\sin (u) }{\cos^{n}(u)} \, du \\ \\ \texttt{By Integration by Parts,} \\ \\ \begin{array}{l | l} \tt u = \sin (u) & \tt dv = \dfrac{\sin (u) }{\cos^{n}(u)}\, du \\ \\ \tt du = \cos (u)\, du & \displaystyle \tt v = \int \dfrac{\sin (u) }{\cos^{n}(u)}\, du = \dfrac{\sec^{n - 1}(u)}{n - 1} \end{array} \\ \\ \\ \displaystyle \tt I_n = \int \sec^{n-2}(u)\, du + \sin (u) \dfrac{\sec^{n - 1} (u)}{n - 1} - \dfrac{1}{n - 1} \int \cos (u) \sec^{n - 1} (u)\, du \\ \\ \displaystyle \tt I_n = \int \sec^{n-2}(u)\, du + \dfrac{\sec^{n - 2}(u)}{n - 1}\cdot \dfrac{\sin (u)}{\cos (u)} - \dfrac{1}{n - 1} \int \sec^{n - 2} (u)\, du \\ \\ \displaystyle \tt I_n = \dfrac{\sec^{n - 2} (u) \tan (u)}{n - 1} + \dfrac{(n - 1) - 1}{n - 1} \int \sec^{n - 2} (u)\, du \\ \\ \displaystyle \tt I_n = \dfrac{\sec^{n - 2} (u) \tan (u)}{n - 1} + \dfrac{n - 2}{n - 1} \int \sec^{n - 2} (u)\, du \\ \\ \\ \therefore \red{\boxed{\displaystyle \tt \int \sec^n(u)\, du=\frac{\sec^{n-2}(u)\tan(u)}{n-1}+\frac{n-2}{n-1}\int \sec^{n-2}(u)\, du}} \\ \\ \bold{Q.E.D.} \end{array}$

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