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NDA 2015
Exam
NDA
Year
2015
Subject
Mathematics
Chapter
Indefinite Integration

NDA 2015 · Mathematics · Indefinite Integration

The integral ∫ (d x)/(a x+b x) is of the form (1)/(r) In [...

The integral ∫dxacos⁡x+bsin⁡x\int \frac{d x}{a \cos x+b \sin x} is of the form 1rIn⁡[tan⁡(x+α2)].Whatis\frac{1}{r} \operatorname{In} \left[\tan \left(\frac{x+\alpha}{2}\right)\right]. What is requalto?equal to?

Options

  1. Aa2+b2a^2+b^2
  2. Ba2+b2\sqrt{a^2+b^2}
  3. Ca+ba+b
  4. Da2−b2\sqrt{a^2-b^2}

Correct answer

B. a2+b2\sqrt{a^2+b^2}

Step-by-step solution

Given that, ∫dxacos⁡x+bsin⁡x=1rln⁡[tan⁡(x+α2)]Let\int \frac{\mathrm{dx}}{\mathrm{a} \cos \mathrm{x}+\mathrm{b} \sin \mathrm{x}}=\frac{1}{\mathrm{r}} \ln \left[\tan\left(\frac{\mathrm{x}+\alpha}{2}\right)\right] Let\mathrm{a}=\mathrm{r}sin⁡α,\sin \alpha,\mathrm{b}=\mathrm{r}cos⁡α\cos \alpha∫dxrsin⁡αcos⁡x+rcos⁡αsin⁡x=1r∫1sin⁡(x+α)=1r∫\ooperatornamecosec(x+α)dx=1rln⁡[tan⁡(x+α2)]\begin{aligned}& \int \frac{d x}{r \sin \alpha \cos x+r \cos \alpha \sin x}=\frac{1}{r} \int \frac{1}{\sin (x+\alpha)}\\ & =\frac{1}{r} \int \ooperatorname{cosec}(x+\alpha)d x=\frac{1}{r} \ln \left[\tan \left(\frac{x+\alpha}{2}\right)\right] \end{aligned}$\begin{aligned}& a=r \sin \alpha \Rightarrowa^2=r^2 \sin ^2 \alpha ...(i)\\ & b=r \cos \alpha \Rightarrowb^2=r^2 \cos ^2 \alpha...(ii) \end{aligned} Adding (i) and (ii), we get ⇒\Rightarrow \quadr2=a2+b2⇒r=a2+b2.\begin{aligned} \mathrm{r}^2& =\mathrm{a}^2+\mathrm{b}^2\\ \Rightarrow \quad \mathrm{r}& =\sqrt{\mathrm{a}^2+\mathrm{b}^2} . \end{aligned}$$
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