Question

Integration of sinx power 6 plus cosx power 6 divided by sin square x into cos square x​

Answer 1

Answer:

$\bold\red{tanx-cotx-3x+C}$

Step-by-step Explanation:

We have to integrate,

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

Using the formula,

$( {a}^{3} + {b}^{3}) = {(a + b)}^{3} - 3ab(a + b)$

We get,

Integration, I

$= \int \: \frac{ {( { \sin }^{2}x + { \cos }^{2} x) }^{ 3} - 3 { \sin }^{2} x{ \cos }^{2} x( { \sin}^{2} x + { \cos }^{2}x) }{ { \sin }^{2}x { \cos }^{2}x } dx \\ \\ = \int \frac{(1 - 3 { \sin }^{2}x { \cos }^{2}x) }{ { \sin }^{2} x { \cos }^{2} x} dx \\ \\ = \int( \frac{1}{ { \sin}^{2}x { \cos }^{2} x } - 3)dx \\ \\ = \int( \frac{ { (\sin}^{2} x + { \cos }^{2}x) }{ { \sin }^{2}x { \cos }^{2} x } - 3)dx \\ \\ = \int( \frac{1}{ { \cos }^{2} x} + \frac{1}{ { \sin }^{2} x} - 3)dx \\ \\ = \int( { \sec }^{2} x + { \cosec }^{2} x - 3)dx \\ \\ = (tanx - cotx - 3x + c)$

Hence,

Integration = $\bold\orange{tanx-cotx-3x+C}$