A) \[-\frac{1}{3}{{(1-{{\tan }^{2}}x)}^{3/2}}+c\]
B) \[\frac{1}{3}{{(1-{{\tan }^{2}}x)}^{3/2}}+c\]
C) \[-\frac{2}{3}{{(1-{{\tan }^{2}}x)}^{2/3}}+c\]
D) None of these
Correct Answer: A
Solution :
\[\int_{{}}^{{}}{\tan x\,.\,{{\sec }^{2}}x\sqrt{1-{{\tan }^{2}}x}\,dx}\] Put \[\tan x=t\Rightarrow {{\sec }^{2}}x\,dx=dt,\] then it reduces to \[\int_{{}}^{{}}{t\sqrt{1-{{t}^{2}}}\,dt}\] Now again, put \[1-{{t}^{2}}=u,\] then its reduced form is \[-\frac{1}{2}\int_{{}}^{{}}{\sqrt{u}\,du}=-\frac{1}{3}{{u}^{3/2}}+c=-\frac{1}{3}{{(1-{{\tan }^{2}}x)}^{3/2}}+c.\]
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