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Question

Mathematics Question on integral

∫dx9x−4x2∫\frac{dx}{\sqrt{9x-4x^2}} equals

A

19sin−1(9x−88)+C\frac{1}{9}sin^{-1}(\frac{9x-8}{8})+C

B

12sin−1(8x−99)+C\frac{1}{2}sin^{-1}(\frac{8x-9}{9})+C

C

13sin−1(9x−88)+C\frac{1}{3}sin^{-1}(\frac{9x-8}{8})+C

D

12sin−1(9x−89)+C\frac{1}{2}sin^{-1}(\frac{9x-8}{9})+C

Answer

12sin−1(8x−99)+C\frac{1}{2}sin^{-1}(\frac{8x-9}{9})+C

Explanation

Solution

The correct answer is B: =12sin−1(8x−99)+C= \frac{1}{2}sin^{-1}(\frac{8x-9}{9})+C
∫dx9x−4x2∫\frac{dx}{\sqrt{9x-4x^2}}
=∫1−4(x2−94x)dx= ∫\frac{1}{\sqrt{-4(x^2-\frac{9}{4x})}}dx
=∫1−4(x2−94x+8164−8164)dx= ∫\frac{1}{-4(x^2-\frac{9}{4}x+\frac{81}{64}-\frac{81}{64})}dx
=∫1−4[(x−98)2−(98)2]dx= ∫\frac{1}{\sqrt{-4[(x-\frac{9}{8})^2-(\frac{9}{8})^2}}] dx
=12∫1(98)2−(x−98)2dx=\frac{1}{2} ∫\frac{1}{\sqrt{(\frac{9}{8})^2 - (x-\frac{9}{8})^2}} dx
=12[sin−1(x−9898)]+C=\frac{1}{2}[sin^{-1}\bigg(\frac{x-\frac{9}{8}}{\frac{9}{8}}\bigg)]+C (∫dya2−y2=sin−1ya+C)( ∫\frac{dy}{\sqrt{a^2-y^2}} = sin^{-1} \frac{y}{a}+C)
=12sin−1(8x−99)+C= \frac{1}{2}sin^{-1}(\frac{8x-9}{9})+C
Hence, the correct Answer is B.