Q112 / 180 · KCET 2015 · Mathematics
\( \int_{-\Pi / 4}^{\text {II } / 4} \frac{d x}{1+\cos 2 x} \) is equal to
- A\( 12 \)
- Correct answerB\( 11 \)
- C\( 04 \)
- D\( 00 \)
Solution
Given that \( I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{d x}{1+\cos 2 x} \)
We know that, \( 1+\cos 2 x=2 \cos ^{2} x \)
So, \( I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{d x}{2 \cos ^{2} x}=\frac{1}{2} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sec ^{2} x d x \)
We know that if \( \mathrm{f}(\mathrm{x}) \) is even function then,
\[ \begin{array}{l} \int_{-a}^{a} f(x) d x=2 \int_{0}^{a} f(x) d x \\ \text { So, } I=\frac{1}{2} \times 2 \int_{0}^{\frac{\pi}{4}} \sec ^{2} x d x=\int_{0}^{\frac{\pi}{4}} \sec ^{2} x d x \\ =[\tan x]_{0}^{\frac{r}{4}}=\tan \frac{\Pi}{4}-\tan 0=1 \end{array} \]
We know that, \( 1+\cos 2 x=2 \cos ^{2} x \)
So, \( I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{d x}{2 \cos ^{2} x}=\frac{1}{2} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sec ^{2} x d x \)
We know that if \( \mathrm{f}(\mathrm{x}) \) is even function then,
\[ \begin{array}{l} \int_{-a}^{a} f(x) d x=2 \int_{0}^{a} f(x) d x \\ \text { So, } I=\frac{1}{2} \times 2 \int_{0}^{\frac{\pi}{4}} \sec ^{2} x d x=\int_{0}^{\frac{\pi}{4}} \sec ^{2} x d x \\ =[\tan x]_{0}^{\frac{r}{4}}=\tan \frac{\Pi}{4}-\tan 0=1 \end{array} \]
Hence, the correct answer is option: 'B'.