Question

The integral
$\int\limits {\frac{\sin ^{2}x \cos ^{2} x }{( \sin ^{5} x + \cos ^{3} x \sin ^{2} x + \sin ^{3} x \cos^{2} x + \cos ^{5} x)^{2} } } \, dx$
Is equal to ?

Answer 1

$\sf \red{We \: have, }$

$\\ I = \int\limits {\frac{\sin ^{2}x \cos ^{2} x }{( \sin ^{5} x + \cos ^{3} x \sin ^{2} x + \sin ^{3} x \cos^{2} x + \cos ^{5} x)^{2} } } \, dx$

$= \int \limits \: \frac{ { \sin}^{2} x. { \cos}^{2}x }{ \Big[ \ { \sin}^{3}x( { \sin}^{2} x \: + { \cos}^{2}x) \: + { \cos}^{3} x( { \sin}^{2}x + { \cos}^{2} x)\Big ]^{2} }dx$

$= \int\limits \: \frac{ { \sin}^{2}x. { \cos}^{2}x }{( { \sin}^{3} x + { \cos}^{3}x)^{2} } dx$

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

$= \int \limits \frac{ { \tan}^{2} x. { \sec}^{2}x }{(1 + { \tan}^{3} x) ^{2} } dx$

$\sf \red{put :} \: { \tan^{3} x \: = \: t} \implies \: 3 { \tan}^{2} x .\: { \sec}^{2} x.dx \: = dt$

$\therefore \: I \: = \: \frac{1}{3} \int \limits \frac{dt}{ {(1 + t)}^{2} }$

$\implies \: I \: = \: \frac{ - 1}{3(1 + t)} + c$

$\boxed{ \boxed{\: I \: = \frac{ - 1}{3(1 + { \tan}^{3} x)} + c }}$

Answer 2

Answer:

$\sf I= \dfrac{-1}{3(tan^3x+1)} +C$

Step-by-step explanation:

Given:

$\sf \dfrac{sin^2x\:cos^2x}{(sin^5x+cos^3xsin^2x+sin^3xcos^2x+cos^5x)^2}$

To Find:

$\displaystyle \sf \int\limits { \dfrac{sin^2x\:cos^2x}{(sin^5x+cos^3xsin^2x+sin^3xcos^2x+cos^5x)^2}} \, dx$

Solution:

$\displaystyle \sf \int\limits { \dfrac{sin^2x\:cos^2x}{(sin^5x+cos^3xsin^2x+sin^3xcos^2x+cos^5x)^2}} \, dx$

Taking sin²x and cos²x common from the denominator,

$\displaystyle \sf \int\limits { \dfrac{sin^2x\:cos^2x}{(sin^2x(sin^3x+cos^3x)+cos^2x(sin^3x+cos^3x))^2}} \, dx$

Now taking sin³x + cos³x as common,

$\displaystyle \sf \int\limits {\dfrac{sin^2x\:cos^2x}{((sin^3x+cos^3x)(sin^2x+cos^2x))^2} } \, dx$

We know that,

sin²x + cos²x = 1

Hence,

$\displaystyle \sf \int\limits {\dfrac{(sin^2x\:cos^2x)}{(sin^3x+cos^3x)^2} } \, dx$

Taking cos³ x as common,

$\displaystyle \sf \int\limits {\dfrac{(sin^2x\:cos^2x)}{cos^6x(tan^3x+1)^2} } \, dx$

$\displaystyle \sf \int\limits {\dfrac{sin^2x}{cos^4x(tan^3x+1)^2} } \, dx$

$\displaystyle \sf \int\limits {\dfrac{sin^2x}{cos^2x\times cos^2x(tan^3x+1)^2} } \, dx$

$\displaystyle \sf \int\limits {\dfrac{tan^2x\times sec^2x }{(tan^3x+1)^2} } \, dx$

Now let us assume t = tan³x + 1

Differentiating on both sides,

We know

$\sf \dfrac{d}{dx} (tan\:x)=sec^2\:x$

dt = 3 tan²x × sec²x dx

dt/3 = tan²x sec²x dx

Substituting this above

$\displaystyle \sf \int\limits {\dfrac{dt}{3t^2} } \,$

Taking 1/3 outside,

$\displaystyle \sf \dfrac{1}{3}\times \int\limits {\dfrac{dt}{t^2} } \,$

$\sf \implies \dfrac{1}{3} \times \dfrac{t^{-2+1}}{-2+1} +C$

$\sf \implies \dfrac{1}{3} \times \dfrac{t^{-1}}{-1} +C$

$\sf \implies \dfrac{-1}{3} \times t^{-1} +C$

Give back the value of t,

$\sf \implies \dfrac{-1}{3} \times (tan^3x+1)^{-1} +C$

$\sf \implies \dfrac{-1}{3(tan^3x+1)} +C$