Qus : 41
2
The value of the integral $\displaystyle\int_{\pi/4}^{3\pi/4}\frac{x}{1+\sin x},dx$ is :
3
$\dfrac{\pi}{2}(\sqrt{2}+1)$
✓ Solution
Qus : 42
3
The value of $\int\limits_0^\pi {{{{e^{\cos x}}\sin x} \over {(1 + {{\cos }^2}x)({e^{\cos x}} + {e^{ - \cos x}})}}dx} $ is equal to:
✓ Solution
Qus : 43
1
If ${I_n} = \int\limits_{{\pi \over 4}}^{{\pi \over 2}} {{{\cot }^n}x\,dx} $, then :
1
${1 \over {{I_2} + {I_4}}},{1 \over {{I_3} + {I_5}}},{1 \over {{I_4} + {I_6}}}$ are in A.P.
2
I2 + I4 , I3 + I5 , I4 + I6 are in A.P.
3
${1 \over {{I_2} + {I_4}}},{1 \over {{I_3} + {I_5}}},{1 \over {{I_4} + {I_6}}}$
4
I2 + I4 , (I3 + I5 )2 , I4 + I6 are in G.P.
✓ Solution
Qus : 44
1
Let $I_n(x) = \int_0^x \dfrac{1}{(t^2+5)^n} \, dt, \; n = 1, 2, 3, \dots$
Then :
1
$50 I_6 - 9 I_5 = x I_5'$
2
$50 I_6 - 11 I_5 = x I_5'$
3
$50 I_6 - 9 I_5 = I_5'$
4
$50 I_6 - 11 I_5 = I_5'$
✓ Solution
Qus : 45
4
Let $f$ and $g$ be continuous functions on $[0,a]$ such that $f(x)=f(a-x)$ and $g(x)+g(a-x)=4$. Then $\displaystyle \int_{0}^{a} f(x)\,g(x)\,dx$ is equal to :
1
$4\displaystyle\int_{0}^{a} f(x)\,dx$
2
$-3\displaystyle\int_{0}^{a} f(x)\,dx$
3
$\displaystyle\int_{0}^{a} f(x)\,dx$
4
$2\displaystyle\int_{0}^{a} f(x)\,dx$
✓ Solution
Qus : 46
3
$\int\limits_6^{16} {{{{{\log }_e}{x^2}} \over {{{\log }_e}{x^2} + {{\log }_e}({x^2} - 44x + 484)}}dx} $$ is equal to :
✓ Solution
Qus : 47
3
For $0 < \mathrm{a} < 1$, the value of the integral $\int_\limits0^\pi \frac{\mathrm{d} x}{1-2 \mathrm{a} \cos x+\mathrm{a}^2}$ is :
1
$\dfrac{\pi^{2}}{\pi+a^{2}}$
2
$\dfrac{\pi^{2}}{\pi-a^{2}}$
3
$-\dfrac{\pi}{1-a^{2}}$
✓ Solution
Qus : 48
4
The value of $\int\limits_{ - \pi /2}^{\pi /2} {{{{{\cos }^2}x} \over {1 + {3^x}}}} dx$ is :
✓ Solution
Qus : 49
2
The value of $\sum\limits_{n = 1}^{100} {\int\limits_{n - 1}^n {{e^{x - [x]}}dx} } $, where [ x ] is the greatest integer $ \le $ x, is :
✓ Solution
Qus : 50
2
If $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{96 x^2 \cos ^2 x}{\left(1+e^x\right)} \mathrm{d} x=\pi\left(\alpha \pi^2+\beta\right), \alpha, \beta \in \mathbb{Z}$, then $(\alpha+\beta)^2$ equals
✓ Solution