Question

1/2∫ xsin⁻¹ x/√1-x² dx ,Evaluate it.0

Answer 1

Step-by-step explanation:

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Answer 2

$\bold{\int_{0}^{1 / 2} \frac{x \sin ^{-1} x}{\sqrt{1-x^{2}}} d x=\frac{1}{2}-\left(\frac{\pi}{6}\right) \sqrt{\frac{3}{4}}}$

Given:

$\int_{0}^{1 / 2} \frac{x \sin ^{-1} x}{\sqrt{1-x^{2}}} d x$

Step-by-step explanation:

Let us consider the given integral as ‘I’.

$I=\int_{0}^{1 / 2} \frac{x \sin ^{-1} x}{\sqrt{1-x^{2}}} d x$

This is solved by substitution method.

Let $t=\sin ^{-1}(x)$

$\frac{d t}{d x}=\frac{1}{\sqrt{1-x^{2}}}$

$\mathrm{dt}=\frac{d x}{\sqrt{1-x^{2}}}$

Now, the integral becomes,

$I =\int \frac{x \sin ^{-1} x}{\sqrt{1-x^{2}}} d x=\int \sin t \times t \times \frac{d x}{\sqrt{1-x^{2}}}$

The above equation is in trigonometric algebraic form.

$\left[\begin{array}{c}{\left.\int f(x) g(x) d x=f(x) \int g(x) d x=\int\left(f^{\prime}(x) \int g(x) d x\right) d x\right} \\{\Rightarrow f(x)=t} \\{\Rightarrow g(x)=\sin t}\end{array}\right]$

$I=t \int \sin t d t-\int \frac{d(t)}{d t} \int \sin t d t d t$

$I =\mathrm{t}(-\cos \mathrm{t})-\int(-\cos t) d t$

$I =-t \cos t+\int \cos t d t$

$I =-t \cos t+\sin t$

Now,

$t=\sin ^{-1}(x)$

$\sin t=x$

$\sin t=x$

$\sin ^{2} t=x^{2}$

$1-\cos ^{2} t=x^{2}$

$\cos ^{2} t=1-x^{2}$

$\cos t=\sqrt{1-x^{2}}$

On substituting the values, we get,

$I=\int_{0}^{1 / 2} \frac{x \sin ^{-1} x}{\sqrt{1-x^{2}}} d x=\left[x-\sqrt{1-x^{2}} \sin ^{-1} x\right]_{0}^{1 / 2}$

$I=\frac{1}{2}-\sqrt{1-\frac{1}{4}}\left(\sin ^{-1} \frac{1}{2}\right)-0-\sqrt{1-0}\left(\sin ^{-1} 0\right)$

$I=\frac{1}{2}-\sqrt{\frac{3}{4}}\left(\frac{\pi}{6}\right)-0$

$\therefore I=\int_{0}^{1 / 2} \frac{x \sin ^{-1} x}{\sqrt{1-x^{2}}} d x=\frac{1}{2}-\left(\frac{\pi}{6}\right) \sqrt{\frac{3}{4}}$