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Question: Evaluate the integral \[ \int_{0}^{1}{\dfrac{x{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}\text{ dx}\text{.}...

Evaluate the integral ∫01xsin⁡−1x1−x2 dx. \int_{0}^{1}{\dfrac{x{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}\text{ dx}\text{.}}

Explanation

Solution

In this we find the integral ∫01xsin⁡−1x1−x2 dx\int_{0}^{1}{\dfrac{x{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}\text{ dx}} by using methods integration by substitution and integration by parts. In method integration by substitution, we reduce the given function to standard form by using some suitable substitution and the method of integration is used when the integrand can be expressed as a product of two suitable functions, one of which can be differentiable and the other can be integrated.

Complete step-by-step solution:
The following give the rule of integration by parts, if u and v are functions of x then,
∫u⋅v dx=u⋅∫v dx−∫[dudx∫v dx]dx.\int{u\cdot v\text{ d}x=u\cdot \int{v\text{ }}}dx-\int{\left[ \dfrac{du}{dx}\int{v}\text{ }dx \right]}dx.
Where u and v are chosen by the rule of ‘ILATE’.
Let I=∫01xsin⁡−1x1−x2 dx. ...(1)I=\int_{0}^{1}{\dfrac{x{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}\text{ dx}\text{.}}\text{ }...\text{(1)}
Now, we will use the method of integration by substitution to proceed further.
Substitute sin⁡−1x=t ...(2){{\sin }^{-1}}x=t\text{ }...\text{(2)}
Differentiating equation (2) with respect to x, we get
⇒11−x2=dtdx\Rightarrow \dfrac{1}{\sqrt{1-{{x}^{2}}}}=\dfrac{dt}{dx}
⇒11−x2 dx=dt ...(3)\Rightarrow \dfrac{1}{\sqrt{1-{{x}^{2}}}}\text{ }dx=dt\text{ }...\text{(3)}
Also by equation (2)
x=sin⁡t ...(4)x=\sin t\text{ }...\text{(4)}
Again by equation (2), whenx=1⇒t=sin⁡−11⇒t=π2x=1\Rightarrow t={{\sin }^{-1}}1\Rightarrow t=\dfrac{\pi }{2}.
When x=0⇒t=sin⁡−10⇒t=0x=0\Rightarrow t={{\sin }^{-1}}0\Rightarrow t=0
By substituting equation (2), equation (3) and equation (4) in equation (1), we get
I=∫0π2tsin⁡t dt ...(5)I=\int_{0}^{\dfrac{\pi }{2}}{t\sin t\text{ dt}}\text{ }...\text{(5)}
Now we will use integration by parts to solve equation (5),
By we will choose u and v by using rule ILATE,
Since, t is algebraic and sint is trigonometric function
⇒u=t and v=sin⁡t\Rightarrow u=t\text{ and }v=\sin t
By integration by part we get,
I=∫0π2t⋅sin⁡t dt=t⋅∫0π2sin⁡t dt−∫0π2[dtdt∫sin⁡t dt]dt.I=\int_{0}^{\dfrac{\pi }{2}}{t\cdot \sin t\text{ }dt}=t\cdot \int_{0}^{\dfrac{\pi }{2}}{\sin t}\text{ }dt-\int\limits_{0}^{\dfrac{\pi }{2}}{\left[ \dfrac{dt}{dt}\int{\sin t}\text{ }dt \right]d}t.
Since ∫sin⁡t dt=−cos⁡t+c and dtdt=1\int{\sin t\text{ }dt=-\cos t}+c\text{ and }\dfrac{dt}{dt}=1
⇒I=∫0π2t⋅sin⁡t dt=[t⋅(−cos⁡t)]0π2−∫0π2[(1) (−cos⁡t)]dt.\Rightarrow I=\int_{0}^{\dfrac{\pi }{2}}{t\cdot \sin t\text{ }dt}=\left[ t\cdot \left( -\cos t \right) \right]_{0}^{\dfrac{\pi }{2}}-\int\limits_{0}^{\dfrac{\pi }{2}}{\left[ \text{(1) }\left( -\cos t \right) \right]d}t.
⇒I=∫0π2t⋅sin⁡t dt=−[t⋅cos⁡t]0π2+∫0π2cos⁡t dt.\Rightarrow I=\int_{0}^{\dfrac{\pi }{2}}{t\cdot \sin t\text{ }dt}=-\left[ t\cdot \cos t \right]_{0}^{\dfrac{\pi }{2}}+\int\limits_{0}^{\dfrac{\pi }{2}}{\cos t\text{ }d}t.
Since∫cos dt=sin⁡t+c\int{\text{cos }dt=\sin t}+c

& \Rightarrow I=\int_{0}^{\dfrac{\pi }{2}}{t\cdot \sin t\text{ }dt}=-\left[ t\cdot \cos t \right]_{0}^{\dfrac{\pi }{2}}+\left[ \sin t \right]_{0}^{\dfrac{\pi }{2}}. \\\ & \\\ \end{aligned}$$ $$\Rightarrow I=\int_{0}^{\dfrac{\pi }{2}}{t\cdot \sin t\text{ }dt}=\left( -\left[ \dfrac{\pi }{2}\cdot \cos \dfrac{\pi }{2} \right]+\left[ \sin \dfrac{\pi }{2} \right] \right)-\left( -\left[ 0\cdot \cos 0 \right]+\left[ \sin 0 \right] \right).\text{ }...\text{(6)}$$ Since, $\sin \dfrac{\pi }{2}=1,\text{ }\sin 0=0\text{, }\cos \dfrac{\pi }{2}=0\text{ and }\cos 0=1$ By substituting above sine and cosine function in equation (6), we get $$\Rightarrow I=\int_{0}^{\dfrac{\pi }{2}}{t\cdot \sin t\text{ }dt}=\left( -\left[ \dfrac{\pi }{2}\cdot 0 \right]+\left[ 1 \right] \right)-\left( -\left[ 0\cdot 1 \right]+\left[ 0 \right] \right).$$ $$\Rightarrow I=\int_{0}^{\dfrac{\pi }{2}}{t\cdot \sin t\text{ }dt}=\left( 0+\left[ 1 \right] \right)-\left( -\left[ 0 \right]+\left[ 0 \right] \right)=1.$$ $$\Rightarrow I=\int_{0}^{\dfrac{\pi }{2}}{t\cdot \sin t\text{ }dt}=1.$$ $$\Rightarrow I=\int_{0}^{1}{\dfrac{x{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}\text{ dx=1}\text{.}}\text{ }$$ **Note:** In the method of integration by parts one should remember the following points: 1) When the integrand is a product of two functions out of which the second function has to be integrand (whose integration is known). Hence we should make the proper choice of the first function u and second function v. 2) We can choose the first function as the function which comes first in serial order of the letters of the “ILATE” where L stands for logarithmic function, I stands for inverse trigonometric function, A stands for algebraic function, T stands for trigonometric function, E stands for the exponential function. 3) If the integrand contains a logarithmic function or inverse trigonometric function, take it as the first function. In all such cases, if the second function is not given, take it as 1.