Question

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

• A. $\sin x+\cos x+c$
• B. $\tan x+\cot x+c$
• C. $\sec x-\csc x+c$
• D. $\sin x-\cos x+c$

Answer 1

The correct option is C $\sec x-\csc x+c$

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

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

$\therefore {\sin }^{2}x+{\cos }^{2}x=1$

$=\int \frac{{\sin }^{3}x+{\cos }^{3}x}{{\sin }^{2}x{\cos }^{2}x}dx$

$\therefore A^3+B^3=(A+B)(A^2+B^2-AB)$

$=\int \frac{{\sin }^{3}x}{{\sin }^{2}x{\cos }^{2}x}dx+\int \frac{{\cos }^{3}x}{{\sin }^{2}x{\cos }^{2}x}dx$

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

$=\int \sec x\tan xdx+\int cosecx\cot xdx$

$=\sec x-cosecx+c$

Hence this is the answer