Question

sin⁶x+cos⁶x/sin²x cos²x,Integrate the given function w.r.t. x considering them well defined and integrable over proper domain.

Answer 1

we have to find the value of $\int{\frac{sin^6x+cos^6x}{sin^2x.cos^2x}}\,dx$

$sin^6x+cos^6x=(sin^2x)^3+(cos^2x)^3\\=(sin^2x+cos^2x)\{(sin^2x)^2+(cos^2x)^2-sin^2x.cos^2x\}\\=1.\{sin^4x+cos^4-sin^2x.cos^2x\}\\=\{(sin^2x+cos^2x)^2-3sin^2x.cos^2x\}\\=(1-3sin^2x.cos^2x)$

now, $\int{\frac{sin^6x+cos^6x}{sin^2x.cos^2x}}\,dx=\int{\frac{(1-3sin^2x.cos^2x)}{sin^2x.cos^2}}\,dx$

$= \int{\frac{1}{sin^2x.cos^2x}-3}\,dx$

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

$=\int{sec^2x+cosec^2x-3)}\,dx$

$=\int{sec^2x}\,dx+\int{cosec^2x}\,dx-3\int\,dx$

$=tanx-cotx-3x$

hence, answer is tanx - cotx - 3x

Answer 2

In the attachment I have answered this problem.

The integrand is modified in such a way that it is suitable for integration

See the attachment for detailed solution.