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Question

Mathematics Question on Fundamental Theorem of Calculus

∫ax/2a−x−axdx is equal to

A

(A) 1log⁡asin−1⁡(ax)+c

B

(B) 1log⁡atan−1⁡(ax)+c

C

(C) 2a−x−ax+c

D

(D) log⁡(ax−1)+c

Answer

(A) 1log⁡asin−1⁡(ax)+c

Explanation

Solution

Explanation:
Let I=∫dx/2a−x−axdx=∫ax1−a2xdxLet ax=t⇒dxlog⁡adx=dt∴=∫dt1−t2⋅1log⁡a=1log⁡asin−1⁡(t)+c=1log⁡asin−1⁡(ax)+c