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NDA 2014
Exam
NDA
Year
2014
Subject
Mathematics
Chapter
Definite Integration

NDA 2014 · Mathematics · Definite Integration

Paragraph : Consider the integrals I_1=∫_ (π)/(6) ^ (π)/(3) d x 1+...

Paragraph : Consider the integrals I1=∫π6π3dx1+tan⁡x and \mathrm{I}_1=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{d x}{1+\sqrt{\tan x}} \text{ and }\mathrm{I}2=∫π6π3sin⁡xdxsin⁡x+cos⁡x._2=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x} d x}{\sqrt{\sin x}+\sqrt{\cos x}}.Question : What is I1−\mathrm{I}_1-\mathrm{I}2equalto?_2 equal to?

Options

  1. A0
  2. B2 I
  3. Cπ\pi
  4. DNone of the above

Correct answer

A. 0

Step-by-step solution

Consider the integrals I1=∫π/6π/3dx1+tan⁡x\mathrm{I}_1=\int_{\pi / 6}^{\pi / 3} \frac{d x}{1+\sqrt{\tan x}} and \quad\mathrm{I}2=∫π/6π/3sin⁡xdxsin⁡x+cos⁡x_2=\int_{\pi / 6}^{\pi / 3} \frac{\sqrt{\sin x} d x}{\sqrt{\sin x}+\sqrt{\cos x}} Let I1=∫π/6π/3dx1+tan⁡x...(i)=∫π/6π/3cos⁡xcos⁡x+sin⁡xdx=∫π/6π/3cos⁡(π2−x)cos⁡(π2−x)+sin⁡(π2−x)dx[∵(π6+π3−x)=(π2−x)]=∫π/6π/3sin⁡xsin⁡x+cos⁡xdx...(ii)\begin{aligned} \mathrm{I}_1 & =\int_{\pi / 6}^{\pi / 3} \frac{d x}{1+\sqrt{\tan x}} ...(i)\\ & =\int_{\pi / 6}^{\pi / 3} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x \\ & =\int_{\pi / 6}^{\pi / 3} \frac{\sqrt{\cos \left(\frac{\pi}{2}-x\right)}}{\cos \left(\frac{\pi}{2}-x\right)+\sin \left(\frac{\pi}{2}-x\right)} d x \\ & {\left[\because\left(\frac{\pi}{6}+\frac{\pi}{3}-x\right)=\left(\frac{\pi}{2}-x\right)\right] } \\ & =\int_{\pi / 6}^{\pi / 3} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x...(ii) \end{aligned} Adding (i) and (ii) we get 2I1=∫π/6π/31dx=[x]π6π3=π3−π6=π6∴I1=π12\begin{aligned} & 2 \mathrm{I}_1 & =\int_{\pi / 6}^{\pi / 3} 1 d x=[x]_{\frac{\pi}{6}}^{\frac{\pi}{3}}=\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6} \\ \therefore & \mathrm{I}_1 & =\frac{\pi}{12} \end{aligned}I2=∫π/6π/3sin⁡xsin⁡x+cos⁡xdx...(i)I2=∫π/6π/3sin⁡(π2−x)sin⁡(π2−x)+cos⁡(π2−x)dxI2=∫π/6π/3cos⁡xcos⁡x+sin⁡xdx...(ii)\begin{aligned} & \mathrm{I}_2=\int_{\pi / 6}^{\pi / 3} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x ...(i)\\ & \mathrm{I}_2=\int_{\pi / 6}^{\pi / 3} \frac{\sqrt{\sin \left(\frac{\pi}{2}-x\right)}}{\sqrt{\sin \left(\frac{\pi}{2}-x\right)}+\sqrt{\cos \left(\frac{\pi}{2}-x\right)}} d x \\ & \mathrm{I}_2=\int_{\pi / 6}^{\pi / 3} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}} d x...(ii) \end{aligned}Adding(i)and(ii)wegetAdding (i) and (ii) we get2I2=∫π/6π/3sin⁡x+cos⁡xsin⁡x+cos⁡xdx⇒2I2=∫π/6π/31dx=[x]π6π3=π3−π6=π6∴I2=π12.\begin{array}{rlrl} & & 2 \mathrm{I}_2 & =\int_{\pi / 6}^{\pi / 3} \frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x \\ \Rightarrow & 2 \mathrm{I}_2 & =\int_{\pi / 6}^{\pi / 3} 1 d x=[x]_{\frac{\pi}{6}}^{\frac{\pi}{3}}=\frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6} \\ \therefore & \mathrm{I}_2 & =\frac{\pi}{12} . \end{array}I\mathrm{I}}_{1}-\mathrm{I}}_{2}=\frac{\pi}{12}-\frac{\pi}{12}=$0
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