[tex]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} [/tex] About the author Nevaeh
EXPLANATION. [tex]\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} }[/tex] As we know that, We can write equation as, [tex]\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} }[/tex] [tex]\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} }[/tex] As we know that, Formula of : ⇒ sin²x + cos²x = 1. [tex]\sf \implies \displaystyle \int\dfrac{sin^{2} x. cos^{2} x \ dx }{[(sin^{3}x + cos^{3} x)]^{2} }[/tex] [tex]\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} }[/tex] [tex]\sf \implies \displaystyle \int\dfrac{sin^{2} x . cos^{2}x \ dx }{cos^{6}x (tan^{3} x + 1)^{2} }[/tex] [tex]\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} }[/tex] [tex]\sf \implies \displaystyle \int \dfrac{tan^{2} x . sec^{2}x \ dx }{[tan^{3}x + 1 ]^{2} }[/tex] 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. [tex]\sf \implies \displaystyle \dfrac{1}{3} \int \dfrac{dt}{t^{2} }[/tex] [tex]\sf \implies \displaystyle -\bigg(\dfrac{1}{3} \times \dfrac{1}{t} \bigg)+ C.[/tex] Put the value of t = tan³x + 1 in equation, we get. [tex]\sf \implies \displaystyle -\dfrac{1}{3(tan^{3} x + 1)} + C[/tex] 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. Reply
EXPLANATION.
[tex]\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} }[/tex]
As we know that,
We can write equation as,
[tex]\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} }[/tex]
[tex]\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} }[/tex]
As we know that,
Formula of :
⇒ sin²x + cos²x = 1.
[tex]\sf \implies \displaystyle \int\dfrac{sin^{2} x. cos^{2} x \ dx }{[(sin^{3}x + cos^{3} x)]^{2} }[/tex]
[tex]\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} }[/tex]
[tex]\sf \implies \displaystyle \int\dfrac{sin^{2} x . cos^{2}x \ dx }{cos^{6}x (tan^{3} x + 1)^{2} }[/tex]
[tex]\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} }[/tex]
[tex]\sf \implies \displaystyle \int \dfrac{tan^{2} x . sec^{2}x \ dx }{[tan^{3}x + 1 ]^{2} }[/tex]
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.
[tex]\sf \implies \displaystyle \dfrac{1}{3} \int \dfrac{dt}{t^{2} }[/tex]
[tex]\sf \implies \displaystyle -\bigg(\dfrac{1}{3} \times \dfrac{1}{t} \bigg)+ C.[/tex]
Put the value of t = tan³x + 1 in equation, we get.
[tex]\sf \implies \displaystyle -\dfrac{1}{3(tan^{3} x + 1)} + C[/tex]
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.