Integrate the function in Exercise: ^ -1 x - ^ -1 x ^ -1 x + ^ -1 x ,x [0,1]

we know that ^ -1 x + x = 2 ^ -1 x = 2 - ^ -1 x = ^ -1 x ( 2 - ^ -1 x ) 2 dx=2 (2 ^ -1 x - 2 )dx =4 ^ -1 x dx- 1dx=4 ^ -1 x dx-x+c putting x = = ^ 2 =2 = 2 d =4 ( ^ -1 ( ). 2 )d -x+c=4 ( . 2 )d -x+c

Integrate the function in Exercise: ^ -1 x - ^ -1 x ^ -1 x + ^ -1…

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Integrate the function in Exercise:

$\frac{\sin^{-1}\sqrt{\text{x}}-\cos^{-1}\sqrt{\text{x}}}{\sin^{-1}\sqrt{\text{x}}+\cos^{-1}\sqrt{\text{x}}},\text{x}\in$ [0,1]

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Answer

we know that $\sin^{-1}\sqrt{\text{x}}+\cos\sqrt{\text{x}}=\frac{\pi}{2}\Rightarrow\cos^{-1}\sqrt{\text{x}}=\frac{\pi}{2}-\sin^{-1}\sqrt{\text{x}}$
$\therefore\text{I}=\int\frac{\sin^{-1}\sqrt{\text{x}}\Big(\frac{\pi}{2}-\sin^{-1}\sqrt{\text{x}}\Big)}{\frac{\pi}{2}}\text{dx}=\frac{2}{\pi}\int\Big(2\sin^{-1}\sqrt{\text{x}}-\frac{\pi}{2}\Big)\text{dx}$
$\Rightarrow\text{I}=\frac{4}{\pi}\int\sin^{-1}\sqrt{\text{x}}\text{dx}-\int1\text{dx}=\frac{4}{\pi}\int\sin^{-1}\sqrt{\text{x}}\ \text{dx}-\text{x}+\text{c}$
putting $\sqrt{\text{x}}=\sin\theta\Rightarrow\text{x}=\sin^{2}\theta\Rightarrow\text{dx}=2\sin\theta\cos\theta\text{d}\theta=\sin2\theta\ \text{d}\theta$
$\therefore\text{I}=\frac{4}{\pi}\int\big(\sin^{-1}(\sin\theta).\sin2\theta\big)\text{d}\theta-\text{x}+\text{c}=\frac{4}{\pi}\int(\theta.\sin2\theta)\text{d}\theta-\text{x}+\text{c}$

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