Question

Assertion :Consider $I={\int }_{-\frac{\pi }{4}}^{\frac{\pi }{4}}\frac{dx}{1-\sin x}$ $=0$ Reason: ${\int }_{-a}^{a}f\left(x\right)dx=0$, wherever $f\left(x\right)$ is an odd function.

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is not correct but Reason is correct

Answer 1

The correct option is D Assertion is not correct but Reason is correct

$I={\int }_{-\pi /4}^{\pi /4}\frac{dx}{1-\sin x}$
Here, $f\left(x\right)=\frac{1}{1-\sin x}$
f(x) is neither even nor odd.
Now, $I={\int }_{-\pi /4}^{\pi /4}\frac{dx}{1-\sin x}$
$={\int }_{-\pi /4}^{\pi /4}\frac{dx}{1-\frac{2\tan \left(x/2\right)}{1+{\tan }^2\left(x/2\right)}}$

$={\int }_{-\pi /4}^{\pi /4}\frac{{\sec }^2\left(x/2\right)dx}{1+{\tan }^2\left(x/2\right)-2\tan \left(x/2\right)}$

$I={\int }_{-\pi /4}^{\pi /4}\frac{{\sec }^2\left(x/2\right)dx}{{(\tan \left(x/2\right)-1)}^2}$

Put $\tan \left(x/2\right)=t$
$\Rightarrow se{c}^{2}x/2dx=2dt$

$I={\int }_{-\tan \pi /8}^{\tan \pi /8}\frac{2dt}{{(t-1)}^2}$

$I=-2{\left[\frac{1}{(t-1)}\right]}_{-\tan \pi /8}^{\tan \pi /8}$
$\Rightarrow I\ne 0$
Hence, assertion is not true.
Reason is correct statement