A) \[\sqrt{2}.{{\sin }^{-1}}(\operatorname{si}nx-cosx)+c\]
B) \[{{\sin }^{-1}}(\operatorname{si}nx-cosx)+c\]
C) \[\sqrt{2}.{{\cos }^{-1}}(\operatorname{si}nx-cosx)+c\]
D) \[2.{{\cos }^{-1}}(\operatorname{si}nx-cosx)+c\]
Correct Answer: A
Solution :
[a] \[I=\int{(\sqrt{\tan x}+\sqrt{\cot x}).dx}=\int{\left( \frac{\sqrt{\sin x}}{\sqrt{\cos x}}+\frac{\sqrt{\cos x}}{\sqrt{\sin x}} \right)}.dx\] \[=\int{\frac{\sin x+\cos x}{\sqrt{\sin x+\cos x}}.dx}\] \[=\sqrt{2}\int{\frac{\sin x+\cos x}{\sqrt{2\sin x+\cos x}}.dx}=\sqrt{2}\int{\frac{\sin x+\cos x}{\sqrt{1-\left( 1-2\sin x.\cos x \right)}}.dx}\]\[=\sqrt{2}\int{\frac{\sin x+\cos x}{\sqrt{1-{{\left( \sin x-\cos x \right)}^{2}}}}.dx}\] Let \[z=sinx-cosx\] \[\Rightarrow dz=\left( cosx+sinx \right)dx\] So, \[I=\sqrt{2}\int{\frac{dz}{\sqrt{1-{{z}^{2}}}}}=\sqrt{2}.{{\sin }^{-1}}(z)+c\] \[=\sqrt{2}.si{{n}^{-1}}(sinx-cosx)+c\] Hence, option [a] is correct.
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