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The integral $$\int {\sqrt {1 + 2\cot x(\cos ecx + \cot x)\,} \,\,dx} $$ $$\left( {0 < x < {\pi \over 2}} \right)$$ is equal to : (where C is a constant of integration)

JEE · Math · previous-year question

  1. A.4 log(sin $${x \over 2}$$ ) + C
  2. B.2 log(sin $${x \over 2}$$ ) + Ccorrect
  3. C.2 log(cos $${x \over 2}$$ ) + C
  4. D.4 log(cos $${x \over 2}$$) + C

Answer

B. 2 log(sin $${x \over 2}$$ ) + C

Explanation

Let, I = $$\int {\sqrt {1 + 2\cot x\cos ec + 2{{\cot }^2}x} .dx} $$ $$ \Rightarrow $$ I = $$\int {\sqrt {{{{{\sin }^2}x + 2\cos x + 2{{\cos }^2}x} \over {{{\sin }^2}x}}} .dx} $$ $$ \Rightarrow $$ I = $$\int {\sqrt {{{1 + 2\cos x + {{\cos }^2}x} \over {\sin x}}} .dx} $$ $$ \Rightarrow $$ I = $$\int {\left| {{{1 + \cos x} \over {\sin x}}} \right|dx} $$ $$ \Rightarrow $$ I = $$\int {\left| {\cos ec\,x + \cot x} \right|.dx} $$ $$ \Rightarrow $$ I = $$\log \left| {\cos ec\,x - \cot x} \right| + \log \left| {\sin x} \right| + C$$ $$ \Rightarrow $$ I = $$\log \left| {1 - \cos x} \right| + C$$ $$ \Rightarrow $$ I = $$\log \left| {2{{\sin }^2}{x \over 2}} \right| + C$$ $$ \Rightarrow $$ I = $$\log \left| {{{\sin }^2}{x \over 2}} \right| + \log 2+ C$$ $$ \Rightarrow $$ I = 2$$\log \left| {{{\sin }}{x \over 2}} \right| + C_1$$

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