Question

Find the integrals of the function: $\frac{sin^3\ x \ + \ cos^3\ x}{sin^2\ x\ cos^2\ x}$

Answer 1

Answer:

Step-by-step explanation:

Concept:

$\int{secx.tanx}\:dx=secx+c\\\\\int{cosecx.cotx}\:dx= - cosecx +c$

The given integral is solved by decomposition method.

In decomposition method, the given integrand(non-integrable function) is decomposed into integrable functions by using algebraic identities, trigonometric identities, etc.

Now,

$\int{\frac{sin^3x+cos^3x}{sin^2x.cos^2x}}\:dx\\\\=\int[\frac{sin^3x}{sin^2x.cos^2x}+\frac{cos^3x}{sin^2x.cos^2x}]dx\\\\=\int[\frac{sinx}{cos^2x}+\frac{cosx}{sin^2x}]dx\\\\=\int[\frac{1}{cosx}.\frac{sinx}{cosx}+\frac{1}{sinx}.\frac{cosx}{sinx}]dx\\\\=\int[secx.tanx+cosecx.cotx]dx\\\\=\int{secx.tanx}\:dx+\int{cosecx.cotx}\;dx\\\\=secx +(-cosecx)+c\\\\= secx - cosecx +c$