Question

$\int \frac{1}{sinxco{s}^{2}x}=dx$.

• A. $secx+log\mid sec+tan\mid +c$
• B. sex .tanx+c
• C. $secx+log\mid sec-tan\mid +c$
• D. $secx+log\mid cosecx-cotx\mid +c$

Answer 1

The correct option is D $secx+log\mid cosecx-cotx\mid +c$

$\int \frac{1}{\sin x{\cos }^{2}x}dx=\int \frac{{\sin }^{2}x+{\cos }^{2}x}{\sin x{\cos }^{2}x}dx$

$=\int \frac{{\sin }^{2}x}{\sin x{\cos }^{2}x}dx+\int \frac{{\cos }^{2}x}{\sin x{\cos }^{2}x}dx$

$=\int \frac{\sin x}{{\cos }^{2}x}dx+\int \frac{1}{\sin x}dx$

$=\int \sec x\cdot \tan xdx+\int cosecxdx$

$=\sec x+ln|cosecx-\cot x|+c$.