Question

$integrate (\sin^{2}(x) \cos ^{2} (x) ) \div (sin^{5}(x) + \cos^{3}(x) \sin^{2}(x) + \sin^{3}(x) \cos ^{2}(x) + \cos^{5}(x))^{2}$
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Answer 1

EXPLANATION.

$\sf \implies \displaystyle \int \dfrac{sin^{2}x \ cos^{2}x \ dx}{[sin^{5} x + cos^{3} x.sin^{2}x + sin^{3}x.cos^{2} x + cos^{5}x ]^{2} }$

As we know that,

We can write equation as,

$\sf \implies \displaystyle \int \dfrac{sin^{2}x . cos^{2}x \ dx }{[sin^{2}x (sin^{3} x + cos^{3} x )+ cos^{2}x (sin^{3}x + cos^{3} x)]^{2} }$

$\sf \implies \displaystyle \int \dfrac{sin^{2}x. cos^{2}x \ dx }{[(sin^{2}x + cos^{2}x) (sin^{3}x + cos^{3} x)]^{2} }$

As we know that,

Formula of :

⇒ sin²x + cos²x = 1.

$\sf \implies \displaystyle \int\dfrac{sin^{2} x. cos^{2} x \ dx }{[(sin^{3}x + cos^{3} x)]^{2} }$

$\sf \implies \displaystyle \int \dfrac{sin^{2}x . cos^{2}x \ dx }{cos^{6}x \bigg(\dfrac{sin^{3} x}{cos^{3}x }\ + \dfrac{cos^{3}x }{cos^{3}x } \bigg)^{2} }$

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

$\sf \implies \displaystyle \int\dfrac{\bigg(\dfrac{sin^{2}x }{cos^{2}x } \ \times \dfrac{cos^{2}x }{cos^{4}x } \bigg)dx }{[tan^{3}x + 1]^{2} }$

$\sf \implies \displaystyle \int \dfrac{tan^{2} x . sec^{2}x \ dx }{[tan^{3}x + 1 ]^{2} }$

As we know that,

By using substitution method, we get.

Let we assume that,

⇒ tan³x + 1 = t.

Differentiate w.r.t x, we get.

⇒ 3tan²xsec²x dx = dt.

⇒ tan²xsec²x dx = dt/3.

Substitute the value, we get.

$\sf \implies \displaystyle \dfrac{1}{3} \int \dfrac{dt}{t^{2} }$

$\sf \implies \displaystyle -\bigg(\dfrac{1}{3} \times \dfrac{1}{t} \bigg)+ C.$

Put the value of t = tan³x + 1 in equation, we get.

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

                                                                                                                       

MORE INFORMATION.

(1) = If the integration is in the form.

∫eˣ [f(x) + f'(x)]dx = eˣ f(x) + c.

(2) = If the integrals is in the form.

∫ [xf'(x) + f(x)]dx = x f(x) + c.