Question

integrate: (sin^6x +cos^6x) / (sin^2 x.cos ^2 x)

Answer 1

$\int{\frac{sin^6x+cos^6x}{sin^2x.cos^2x}}\,dx$

first resolve , sin^6x + cos^6x
= (sin²x)³ + (cos²x)³

= (sin²x + cos²x)(sin⁴x + cos⁴x - sin²x.cos²x)

=(sin²x + cos²x) {(sin²x + cos²x)² - 3sin²x.cos²x}

but we know, sin²x + cos²x = 1

so, sin^6x + cos^6x = (1 - 3sin²x.cos²x)

or, you can write cos^6x + cos^6x = sin²x + cos²x - 3sin²x.cos²x [ as we know, sin²x + cos²x = 1 ]

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

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

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

= tanx - cotx - 3x + C

hence, $\int{\frac{sin^6x+cos^6x}{sin^2x.cos^2x}}\,dx$

= tanx - cotx - 3x + C

Answer 2

Answer:

The answer to your question is given above....

Thanks