56
MCQ (Single Correct Answer)
$\int_{-\pi / 4}^{\pi / 4} \frac{d x}{1+\cos 2 x}$ is equal to
A
1
B
2
C
3
D
4
57
MCQ (Single Correct Answer)
$\int_0^{\pi / 2} \sqrt{1-\sin 2 x} d x$ is equal to
A
$2 \sqrt{2}$
B
$2(\sqrt{2}+1)$
C
$2$
D
$2(\sqrt{2}-1)$
58
MCQ (Single Correct Answer)
$\int_0^{\pi / 4} \cos x e^{\sin x} d x$ is equal to
A
$e+1$
B
$e-1$
C
$e$
D
$-e$
59
MCQ (Single Correct Answer)
$\int \frac{x+3}{(x+4)^2} e^x d x$ is equal to
A
$\mathrm{e}^x\left(\frac{1}{x+4}\right)+C$
B
$\mathrm{e}^{-x}\left(\frac{1}{x+4}\right)+C$
C
$\mathrm{e}^{-x}\left(\frac{1}{x-4}\right)+C$
D
$\mathrm{e}^{2 x}\left(\frac{1}{x-4}\right)+C$
60
If $\int_0^a \frac{1}{1+4 x^2} d x=\frac{\pi}{8}$, then $a=$ ............. .
Explanation
Let $I=\int_0^a \frac{1}{1+4 x^2} d x=\frac{\pi}{8}$
Now, $\quad \int_0^a \frac{1}{4\left(\frac{1}{4}+x^2\right)} d x=\frac{2}{4}\left[\tan ^{-1} 2 x\right]_0^a$
$=\frac{1}{2} \tan ^{-1} 2 a-0=\pi / 8$
$$\begin{aligned} \frac{1}{2} \tan ^{-1} 2 a & =\frac{\pi}{8} \\ \Rightarrow\quad\tan ^{-1} 2 a & =\pi / 4 \\ \Rightarrow\quad 2 a & =1 \\ \therefore\quad a & =\frac{1}{2} \end{aligned}$$