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  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise Revision Exercise Question 89 Maths Textbook Solution.

Provide Solution For R.D.Sharma Maths Class 12 Chapter 18  Indefinite Integrals Exercise  Revision Exercise Question 89 Maths Textbook Solution.

Answers (1)

Answer:

\dpi{100} -\frac{1}{3} t^{\frac{3}{2}}-\frac{1}{3}\left(1+x-x^{2}\right)^{\frac{3}{2}}+\frac{1}{2}\left[\frac{2 x-1}{4} \sqrt{\frac{5}{4}-\left(x-\frac{1}{2}\right)^{2}}+\frac{\sqrt{5}}{8} \sin ^{-1}\left(\frac{2 x-1}{\sqrt{5}}\right)\right]+c

Hint:

You have to find value of A & B.

Given:

\int x \sqrt{1+x-x^{2}} d x

Solution:

\text { let } x=A\left(\left(\frac{d}{d x}\right)\left(1+x-x^{2}\right)\right)+B

  x=A(1-2 x)+B

x=A+B-2 A x

A + B = 0 ,    -2A = 1 

B=\frac{1}{2} \quad A=-\frac{1}{2}

\mathrm{I}=\int\left[-\frac{1}{2}(1-2 x)+\frac{1}{2}\right) \sqrt{1+x-x^{2}} d x

=-\frac{1}{2} \int(1-2 x) \sqrt{1+x-x^{2}} d x+\frac{1}{2} \sqrt{1+x-x^{2}} d x

I=-\frac{1}{2} \int \sqrt{t} d t \ldots .\left(\sqrt{1+x-x^{2}}=\mathrm{t}\right)

=-\frac{1}{3} t^{\frac{3}{2}}=-\frac{1}{3}\left(1+x-x^{2}\right)^{\frac{3}{2}}+c

I I=\frac{1}{2} \int \sqrt{1+x-x^{2}} d x

=\frac{1}{2} \int \sqrt{\frac{5}{4}-\left(x-\frac{1}{2}\right)^{2}} d x                                                    \left[\because x-\frac{1}{2}=z, d x=d z\right]

=\frac{1}{2} \int \sqrt{\left(\frac{\sqrt{5}}{2}\right)^{2}-(z)^{2}} d z

=\frac{1}{2}\left[\frac{1}{2} z \sqrt{\frac{\sqrt{5}}{4}-z^{2}}+\frac{1}{2} \cdot \frac{\sqrt{5}}{4} \sin ^{-1}\left(\frac{2 z}{\sqrt{5}}\right)+c\right]

=\frac{1}{2}\left[\frac{2 x-1}{4} \sqrt{\frac{5}{4}-\left(x-\frac{1}{2}\right)^{2}}+\frac{\sqrt{5}}{8} \sin ^{-1}\left(\frac{2 x-1}{\sqrt{5}}\right)+c\right]

\int x \sqrt{1+x-x^{2}}=I+I I

I=-\frac{1}{3} t^{\frac{3}{2}}-\frac{1}{3}\left(1+x-x^{2}\right)^{\frac{3}{2}}+c+\frac{1}{2}\left[\frac{2 x-1}{4} \sqrt{\frac{5}{4}-\left(x-\frac{1}{2}\right)^{2}}+\frac{\sqrt{5}}{8} \sin ^{-1}\left(\frac{2 x-1}{\sqrt{5}}\right)+c\right]

I=-\frac{1}{3} t^{\frac{3}{2}}-\frac{1}{3}\left(1+x-x^{2}\right)^{\frac{3}{2}}+\frac{1}{2}\left[\frac{2 x-1}{4} \sqrt{\frac{5}{4}-\left(x-\frac{1}{2}\right)^{2}}+\frac{\sqrt{5}}{8} \sin ^{-1}\left(\frac{2 x-1}{\sqrt{5}}\right)\right]+C

 

Posted by

infoexpert21

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