Question

definite integrals....​

Answer 1

Given Integrand,

$\displaystyle \sf \int_{0}^{1} \dfrac{cos {}^{ - 1}x }{ \sqrt{1 - {x}^{2} } } dx$

Let t = arccos(x)

Differentiating w.r.t x on both sides,

$\implies \sf \: \dfrac{dt}{dx} = \dfrac{ - 1}{ \sqrt{1 - {x}^{2} } } \\ \\ \implies \sf \: - dt( \sqrt{1 - {x}^{2} } ) = dx$

When x = 1, t = arccos (x) = 0

When x = 0, t = arccos(x) = π/2

Therefore,

$\longrightarrow \displaystyle \sf - \int_{ \frac{\pi}{2} }^{0} \dfrac{t}{ \sqrt{1 - {x}^{2} } } ( \sqrt{1 - {x}^{2} })dt \\ \\ \longrightarrow \displaystyle \sf - \int_{ \frac{\pi}{2} }^{0} t \: dt \\ \\ \longrightarrow \displaystyle \sf - \dfrac{ \: {t}^{2} }{2} \bigg| _{ \frac{\pi}{2} }^{0} \\ \\ \longrightarrow \displaystyle \sf - \dfrac{ \: {0}^{2} }{2} + \dfrac{ (\frac{\pi}{2}) {}^{2} }{2} \\ \\ \longrightarrow \displaystyle \sf \: \dfrac{ \: \: {\pi}^{2} }{8}$