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Question: \(\int_{}^{}{\frac{\mathbf{5}\mathbf{-}\mathbf{2x}}{\sqrt{\mathbf{6 + x}\mathbf{-}\mathbf{x}^{\mathb...

∫5−2x6+x−x2dx\int_{}^{}{\frac{\mathbf{5}\mathbf{-}\mathbf{2x}}{\sqrt{\mathbf{6 + x}\mathbf{-}\mathbf{x}^{\mathbf{2}}}}\mathbf{dx}} =

A

26+x−x2−4sin⁡−1(2x−15)+c2\sqrt{6 + x - x^{2}} - 4\sin^{- 1}\left( \frac{2x - 1}{5} \right) + c

B

26+x−x2+4sin⁡−1(2x−15)+c2\sqrt{6 + x - x^{2}} + 4\sin^{- 1}\left( \frac{2x - 1}{5} \right) + c

C

−26+x−x2−4sin⁡−1(2x−15)+c- 2\sqrt{6 + x - x^{2}} - 4\sin^{- 1}\left( \frac{2x - 1}{5} \right) + c

D

−26+x−x2+4sin⁡−1(2x−15)+c- 2\sqrt{6 + x - x^{2}} + 4\sin^{- 1}\left( \frac{2x - 1}{5} \right) + c

Answer

26+x−x2+4sin⁡−1(2x−15)+c2\sqrt{6 + x - x^{2}} + 4\sin^{- 1}\left( \frac{2x - 1}{5} \right) + c

Explanation

Solution

I=∫5−2x6+x−x2dxI = \int_{}^{}{\frac{5 - 2x}{\sqrt{6 + x - x^{2}}}dx}

Let 5−2x=Mddx(6+x−x2)+N5 - 2x = M\frac{d}{dx}(6 + x - x^{2}) + N ⇒5−2x=M(1−2x)+N\Rightarrow 5 - 2x = M(1 - 2x) + N

Equating the coefficients of x and constant terms on both sides, we get

−2=−2M⇒M=1- 2 = - 2M \Rightarrow M = 1 and 5=M+N⇒N=5−1=45 = M + N \Rightarrow N = 5 - 1 = 4

∴5−2x=(1−2x)+4\therefore 5 - 2x = (1 - 2x) + 4

Hence,

I=∫(1−2x)+46+x−x2dxI = \int_{}^{}{\frac{(1 - 2x) + 4}{\sqrt{6 + x - x^{2}}}dx} =∫1−2x6+x−x2dx+4∫dx6+x−x2= \int_{}^{}{\frac{1 - 2x}{\sqrt{6 + x - x^{2}}}dx + 4\int_{}^{}\frac{dx}{\sqrt{6 + x - x^{2}}}} =I1+4I2,= I_{1} + 4I_{2}, (say)

Now, I1=∫1−2x6+x−x2dxI_{1} = \int_{}^{}{\frac{1 - 2x}{\sqrt{6 + x - x^{2}}}dx}

Putting 6+x−x2=t⇒(1−2x)dx=dt,6 + x - x^{2} = t \Rightarrow (1 - 2x)dx = dt, we have,

I1=∫dtt=2t+C1=26+x−x2+C1I_{1} = \int_{}^{}{\frac{dt}{\sqrt{t}} = 2\sqrt{t} + C_{1} = 2\sqrt{6 + x - x^{2}} + C_{1}}

andI2=∫dx6+x−x2=∫dx6−(x2−x)I_{2} = \int_{}^{}{\frac{dx}{\sqrt{6 + x - x^{2}}} = \int_{}^{}\frac{dx}{\sqrt{6 - (x^{2} - x)}}}

I=∫dx6−(x2−x+14)+14=∫dx254−(x−12)2=∫du(52)2−u2I = \int_{}^{}{\frac{dx}{\sqrt{6 - \left( x^{2} - x + \frac{1}{4} \right) + \frac{1}{4}}} = \int_{}^{}\frac{dx}{\sqrt{\frac{25}{4} - \left( x - \frac{1}{2} \right)^{2}}}} = \int_{}^{}\frac{du}{\sqrt{\left( \frac{5}{2} \right)^{2} - u^{2}}} (where, u=x−12)\left( \text{where, }u = x - \frac{1}{2} \right)

=sin⁡−1(u5/2)+C2=sin⁡−1[25(x−12)]+C2=sin⁡−1(2x−15)+C2∴I=I1+4I2=26+x−x2+4sin⁡−1(2x−15)+C= \sin^{- 1}\left( \frac{u}{5/2} \right) + C_{2} = \sin^{- 1}\left\lbrack \frac{2}{5}\left( x - \frac{1}{2} \right) \right\rbrack + C_{2} = \sin^{- 1}\left( \frac{2x - 1}{5} \right) + C_{2}\therefore I = I_{1} + 4I_{2} = 2\sqrt{6 + x - x^{2}} + 4\sin^{- 1}\left( \frac{2x - 1}{5} \right) + C (where, C=C1+4C2)(\text{where, C} = C_{1} + 4C_{2})