Question

Evaluate: $\int {\sec }^{4}x.{cosec}^{2}xdx$

• A. $\frac{1}{3}{t}^3+t.$
• B. $\frac{1}{3}{t}^3+2t-\frac{1}{t}.$
• C. $\frac{1}{2}{t}^3+2t-\frac{1}{t}.$
• D. $\frac{1}{3}{t}^3-t-\frac{1}{t}.$

Answer 1

The correct option is B $\frac{1}{3}{t}^3+2t-\frac{1}{t}.$

$I=\int \frac{1}{{\cos }^{4}x{\sin }^{2}x}dx=\int \frac{{({\sin }^{2}x+{\cos }^{2}x)}^2}{{\cos }^{4}x{\sin }^{2}x}dx$ $=\int [\frac{{\sin }^{2}x}{{\cos }^{4}x}+\frac{1}{{\sin }^{2}x}+\frac{2}{{\cos }^{2}x}]dx$ $=\int ({\sec }^{2}x{\tan }^{2}x+co{\sec }^{2}x+2{\sec }^{2}x)dx=\frac{1}{3}{\tan }^{3}x-\cot x+2\tan x$ Alt. Divide above and below by ${\cos }^{4+2}x={\cos }^{6}x$ $\therefore =\int \frac{{\sec }^{4}x,{\sec }^{2}x}{{\tan }^{2}x}dx=\int \frac{{(1+{\tan }^{2}x)}^2}{{\tan }^{2}x}{\sec }^{2}xdx$ $=\int \frac{{(1+t^2)}^2}{t^2}dt=\int (t^2+2+\frac{1}{t^2})dt$ $=\frac{1}{3}{t}^3+2t-\frac{1}{t}etc.$