Evaluate-0-3-4-2xsin-1-2x-1-4x-2-dx-

Question Number 48955 by Abdulhafeez Abu qatada last updated on 30/Nov/18

E v a l u a t e \underset{0}{\int^{\frac{\sqrt{3}}{4}}} \frac{2 x \sin^{- 1} (2 x)}{\sqrt{1 - 4 x^{2}}} d x

Answered by tanmay.chaudhury50@gmail.com last updated on 30/Nov/18

2 x = s i n θ 2 d x = c o s θ d θ

\int_{0}^{\frac{π}{3}} \frac{s i n θ \times θ}{\sqrt{1 - {s i n}^{2} θ}} \times \frac{c o s θ d θ}{2}

\frac{1}{2} \int_{0}^{\frac{π}{3}} θ s i n θ d θ

I = \int θ s i n θ d θ

= θ \int s i n θ d θ - \int [\frac{d θ}{d θ} \int s i n θ d θ] d θ

= - θ c o s θ - \int - c o s θ d θ

= - θ c o s θ + s i n θ + c

s o r e q u i r e d i n t r e g a l i s

\frac{1}{2} ∣ - θ c o s θ + s i n θ ∣_{0}^{\frac{π}{3}}

= \frac{1}{2} [(- \frac{π}{3} \times \frac{1}{2} + \frac{\sqrt{3}}{2})]

= \frac{- π}{12} + \frac{\sqrt{3}}{4}

Answered by MJS last updated on 30/Nov/18

u^{'} = \frac{2 x}{\sqrt{1 - 4 x^{2}}} \Rightarrow u = - \frac{\sqrt{1 - 4 x^{2}}}{4}

v = \arcsin 2 x \Rightarrow v^{'} = \frac{2}{\sqrt{1 - x^{2}}}

\int u^{'} v = u v - \int {u v}^{'}

\int \frac{2 x \arcsin 2 x}{\sqrt{1 - 4 x^{2}}} = x - \frac{\sqrt{1 - 4 x^{2}}}{2} \arcsin 2 x

\underset{0}{\int^{\frac{\sqrt{3}}{4}}} \frac{2 x \arcsin 2 x}{\sqrt{1 - 4 x^{2}}} d x = \frac{3 \sqrt{3} - π}{12}

Answered by Abdulhafeez Abu qatada last updated on 02/Dec/18