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NCERT Class XII Mathematics Chapter - - Solutions

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..... Miscellaneous Exercise .....
Integrate the functions in Exercises 229 to 252.
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Question : 231 of 271
Marks: +1, -0
1xax−x2\frac{1}{x\sqrt{ax - x^2}}
[Hint : Put x=atx = \frac{a}{t}]
Solution:  
Let I=∫1xax−x2 dxI = \int \frac{1}{x\sqrt{ax - x^2}} \, dx
Put x=at⇒dx=−at2 dtx = \frac{a}{t} \Rightarrow dx = -\frac{a}{t^2} \, dt
Now, xax−x2x\sqrt{ax - x^2}
=ata2t−a2t2= \frac{a}{t} \sqrt{\frac{a^2}{t} - \frac{a^2}{t^2}}
=a2t1t−1t2= \frac{a^2}{t} \sqrt{\frac{1}{t} - \frac{1}{t^2}}
=a2t2t−1= \frac{a^2}{t^2} \sqrt{t - 1}
∴I=∫1a2t2t−1×(−at2) dt\therefore I = \int \frac{1}{\frac{a^2}{t^2} \sqrt{t - 1}} \times \left( -\frac{a}{t^2} \right) \, dt
=−1a∫1t−1 dt= -\frac{1}{a} \int \frac{1}{\sqrt{t - 1}} \, dt
=−1a⋅(t−1)−12+1−12+1+C= -\frac{1}{a} \cdot \frac{(t - 1)^{-\frac{1}{2} + 1}}{-\frac{1}{2} + 1} + C
=−2at−1+C= -\frac{2}{a} \sqrt{t - 1} + C
=−2aax−1+C= -\frac{2}{a} \sqrt{\frac{a}{x} - 1} + C
=−2aa−xx+C= -\frac{2}{a} \sqrt{\frac{a - x}{x}} + C
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