Q220 / 270 · NDA 2015 (Phase 2) · Mathematics

The value of $\int_a^b \frac{x^7+\sin x}{\cos x} d x$ where $a+b=0$ is

  • A
    $2 \mathrm{~b}-\mathrm{a} \sin (\mathrm{b}-\mathrm{a})$
  • B
    $a+3 b \cos (b-a)$
  • C
    $\sin \mathrm{a}-(\mathrm{b}-\mathrm{a}) \cos \mathrm{b}$
  • Correct answer
    D
    0

Solution

$\begin{aligned} & a+b=0 \Rightarrow a=-b \\ & I=\int_{-b}^b \frac{x^7+\sin x}{\cos x} \end{aligned}$ Using property $\begin{aligned} & \int_{-a}^a f(x) d x= \begin{cases}\int_0^{2 a} f(x) d x & ; \text { if } f(x) \text { is even } \\ 0 & ; \text { if } f(x) \text { is odd }\end{cases} \\ & f(x)=\frac{x^7+\sin x}{\cos x} \\ & f(-x)=\frac{(-x)^7+\sin (-x)}{\cos (-x)}=\frac{-x^7-\sin x}{\cos x} \\ & =-\left[\frac{x^7+\sin x}{\cos x}\right] \end{aligned}$ $=-\mathrm{f}(\mathrm{x})$ So $f(x)$ is odd hence $\mathrm{I}=0$

Hence, the correct answer is option: 'D'.

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