A) \[\frac{1}{2}{{\sin }^{-1}}({{\sin }^{2}}x)+C\]
B) \[\frac{1}{2}{{\cos }^{-1}}({{\sin }^{2}}x)+C\]
C) \[{{\tan }^{-1}}({{\sin }^{2}}x)+C\]
D) \[{{\tan }^{-1}}(2{{\sin }^{2}}x)+C\]
Correct Answer: A
Solution :
Let \[I=\int{\frac{\sin x\cos x}{\sqrt{1-{{\sin }^{4}}x}}dx}\] Put \[{{\sin }^{2}}x=t\Rightarrow 2\sin \,x\,\cos \,x\,dx=dt\] \[\therefore \] \[I=\int{\frac{dt}{2\sqrt{1-{{t}^{2}}}}}\] \[=\frac{1}{2}{{\sin }^{-1}}t+C\] \[=\frac{1}{2}{{\sin }^{-1}}({{\sin }^{2}}x)+C\]
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