Question

Evaluate: $\int {\sec }^{5}x{\text{cosec}}^{3}xdx$

• A. $\frac{t^4}{4}-\frac{t^{-2}}{-2}+3\frac{t^2}{2}+3\log t+c,wheret={\tan }^{1/2}x$
• B. $\frac{t^4}{4}+\frac{t^{-2}}{-2}+3\frac{t^2}{2}+3\log t+c,wheret={\tan }^{1/2}x$
• C. $\frac{t^4}{4}+\frac{t^{-2}}{-2}-3\frac{t^2}{2}+3\log t+c,wheret={\tan }^{1/2}x$
• D. $\frac{t^4}{4}+\frac{t^{-2}}{-2}-3\frac{t^2}{2}-3\log t+c,wheret={\tan }^{1/2}x$

Answer 1

The correct option is B $\frac{t^4}{4}+\frac{t^{-2}}{-2}+3\frac{t^2}{2}+3\log t+c,wheret={\tan }^{1/2}x$

$\int se{c}^{5}x{\text{cosec}}^{3}xdx$

$=\int \frac{dx}{{\cos }^{5}x{\sin }^{3}x}$

$=\int \frac{dx}{{\cos }^{8}x{\tan }^{3}x}$

$=\int \frac{{\sec }^{6}x{\sec }^{2}xdx}{{\tan }^{3}x}$

$=\int \frac{{(1+{\tan }^{2}x)}^{3}se{c}^{2}xdx}{{\tan }^{3}x}$

$=\int \frac{{(1+t^2)}^{3}dt}{t^3}$

$=\int {(t+\frac{1}{t})}^{3}dt$

$=\int (t^3+\frac{1}{t^3}+3t+\frac{3}{t})dt$

$=\frac{t^4}{4}+\frac{t^{-2}}{-2}+3\frac{t^2}{2}+3\log t+c,wheret={\tan }^{1/2}x$