Question

Solve this integral asap
Don't spam ​

Answer 1

EXPLANATION.

$\sf \implies \int \bigg(\dfrac{Sin^{-1}\sqrt{x} - Cos^{-1}\sqrt{x} }{Sin^{-1}\sqrt{x} + Cos^{-1} \sqrt{x} } \bigg) dx.$

As we know that,

⇒ Sin⁻¹x + Cos⁻¹x = π/2.

⇒ Cos⁻¹x = π/2 - Sin⁻¹x.

Apply this formula we get,

$\sf \implies \int \dfrac{Sin^{-1}\sqrt{x} - \bigg(\dfrac{\pi}{2}- Sin^{-1} \sqrt{x} \bigg) }{\dfrac{\pi}{2} } dx.$

$\sf \implies \int \dfrac{2}{\pi} \bigg(Sin^{-1} \sqrt{x} - \dfrac{\pi}{2} + Sin^{-1}\sqrt{x} \bigg) dx$

$\sf \implies \int \dfrac{2}{\pi} \bigg(2Sin^{-1}\sqrt{x} - \dfrac{\pi}{2} \bigg)dx$

$\sf \implies \int \bigg(\dfrac{4}{\pi} Sin^{-1} \sqrt{x} - 1 \bigg)dx$

$\sf \implies \dfrac{4}{\pi} \int Sin^{-1}\sqrt{x} dx - \int dx.$

$\sf \implies \dfrac{4}{\pi} \int Sin^{-1}\sqrt{x} dx - x + c.$

Let We assume that,

⇒ I₁ = ∫Sin⁻¹√x + c.

⇒ I = 4/π (I₁) - x + c.

$\sf \implies \int Sin^{-1} \sqrt{x} dx$

Let we assume that,

⇒ √x = t.

⇒ x = t².

Differentiate w.r.t x , we get.

⇒ dx = 2tdt.

$\sf \implies \int Sin^{-1} t.2tdt$

2 is a constant term, it come outside

$\sf \implies 2 \int Sin^{-1} t.tdt$

By applying integration by parts we get.

Formula of integration by parts,

∫(u.v)dx = u(∫vdx) - ∫(du/dx). (∫vdx) dx.

By applying ILATE formula.

I = Inverse Trigonometric Function.

L = Logarithmic Function.

A = Algebraic Function.

T = Trigonometric Function.

E = Exponential Function.

t = Algebraic Function.

Sin⁻¹t = Inverse trigonometric functions.

As we know that,

Sin⁻¹x considered as first function.

t considered as second function.

$\sf \implies 2\bigg[Sin^{-1}t \int tdt - \int \bigg((d(sin^{-1} t)/dt . \int tdt )\bigg)dx\bigg]$

$\sf \implies 2 \bigg[ Sin^{-1} t.\dfrac{t^{2} }{2} - \int \dfrac{1}{\sqrt{1 - t^{2} } } \ X \ \dfrac{t^{2} }{2} dt + c \bigg]$

$\sf \implies t^{2} Sin^{-1} t - \int \dfrac{t^{2} }{\sqrt{1 - t^{2} } } dt + c$

$\sf \implies t^{2} Sin^{-1} t + \int \dfrac{-t^{2} }{\sqrt{1 - t^{2} } } dt + c.$

Again integrate the Equation,

⇒ ∫-t²/√1 - t² dt. we get.

$\sf \implies \int \dfrac{-t^{2} }{\sqrt{1 - t^{2} } } dt$

Adding and subtracting numerator by 1.

$\sf \implies \int \dfrac{-t^{2}+ 1 - 1 }{\sqrt{1 - t^{2} } }dt.$

$\sf \implies \int \dfrac{1 - t^{2} }{\sqrt{1 - t^{2} } } dt - \int \dfrac{1}{\sqrt{1 - t^{2} } }dt.$

$\sf \implies \int \sqrt{1 - t^{2} } dt. - \int \dfrac{1}{\sqrt{1 - t^{2} } }dt.$

As we know that,

⇒ ∫√a² - x² dx = x/2 √a² - x² + a²/2 Sin⁻¹(x/a) + c.

⇒ ∫ 1/√a² - x² dx = Sin⁻¹(x/a) + c.

By applying this Formula in equation, we get.

$\sf \implies \dfrac{t}{2} \sqrt{1 - t^{2} } + \dfrac{1}{2}Sin^{-1} t - Sin^{-1} t.$

$\sf \implies \dfrac{t}{2} \sqrt{1 - t^{2} } - \dfrac{1}{2} Sin^{-1}t$

$\sf \implies t^{2}Sin^{-1}t + \dfrac{t}{2} \sqrt{1 - t^{2} } - \dfrac{1}{2}Sin^{-1}t.$

Put the value of t = √x in equation, we get.

$\sf \implies (\sqrt{x} )^{2} Sin^{-1} \sqrt{x} + \dfrac{\sqrt{x} }{2} \sqrt{1 - (\sqrt{x} )^{2}} - \dfrac{1}{2} Sin^{-1}\sqrt{x}$

$\sf \implies x Sin^{-1} \sqrt{x} + \dfrac{\sqrt{x} }{2} \sqrt{1 - x} - \dfrac{1}{2}Sin^{-1} \sqrt{x}$

Put the value in main equation, we get.

$\sf \implies I = \dfrac{4}{\pi} - x + c.$

$\sf \implies \dfrac{4}{\pi} \bigg(x Sin^{-1}\sqrt{x} + \dfrac{\sqrt{x} }{2} \sqrt{1 - x} - \dfrac{1}{2}Sin^{-1} \sqrt{x} \bigg) - x + c$

$\sf \implies I = \dfrac{4}{\pi} \bigg( x Sin^{-1}\sqrt{x} + \dfrac{\sqrt{x} - x^{2} }{2} - \dfrac{1}{2} Sin^{-1} \sqrt{x} \bigg) - x + c$

$\sf \implies I = \dfrac{4}{\pi} x Sin^{-1}\sqrt{x} - \dfrac{2}{\pi} Sin^{-1}\sqrt{x} + \dfrac{2}{\pi} \sqrt{x - x^{2} } - x + c$

$\sf \implies I = Sin^{-1}\sqrt{x} \bigg[\dfrac{4x}{\pi} - \dfrac{2}{\pi} \bigg] + \dfrac{2\sqrt{x - x^{2} } }{\pi} - x + c$

$\sf \implies \boxed{I = Sin^{-1} \sqrt{x} \bigg[\dfrac{2(2x - 1)}{\pi} \bigg] + \dfrac{2\sqrt{x- x^{2} } }{\pi} - x + c.}$

Answer 2

Answer:

hey bro oa dont know this answer