MCQ
$\int\limits_0^\pi {{e^{{{\cos }^4}x}}} . \cos^5(2n + 1)x \,dx, (n \in I)$ is equal to
- A$\pi$
- B$1$
- C$\pi/2$
- ✓$0$
✓
Answer
Correct option: D.
$0$
d
By prop, $-\,7$
By prop, $-\,7$
Ans is zero
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
If $A.M$ and $G.M$ of $x$ and $y$ are in the ratio $p : q$, then $x : y$ is
→$\lim _{n \rightarrow \infty}\left(\frac{n^{2}}{\left(n^{2}+1\right)(n+1)}+\frac{n^{2}}{\left(n^{2}+4\right)(n+2)}+\frac{n^{2}}{\left(n^{2}+9\right)(n+3)}+\ldots+\frac{n^{2}}{\left(n^{2}+n^{2}\right)(n+n)}\right)$ is equal to
→The missing value in the following figure is
If $f(x) = \left\{ \begin{array}{l}\,\,\,\,\,\,\,x,\;{\rm{when\,\, }}0 \le x \le 1\\2 - x,\;{\rm{when \,\,}}1 < x \le 2\end{array} \right.$, then $\mathop {\lim }\limits_{x \to 1} f(x) = $
→If the sum of first $n$ terms of an $A.P.$ is $cn(n -1)$ , where $c \neq 0$ , then sum of the squares of these terms is
→If the value of the integral $\int_{-1}^1 \frac{\cos \alpha x}{1+3^x} d x$ is $\frac{2}{\pi}$. Then, a value of $\alpha$ is
→If $y = {\tan ^{ - 1}}\left( {\frac{1}{{{x^2} + x + 1}}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{{{x^2} + 3x + 3}}} \right) $ $+ {\tan ^{ - 1}}\left( {\frac{1}{{{x^2} + 5x + 7}}} \right) + ......$ up to $n$ terms, then $\frac{dy}{dx}$ is equal to
→If $z = x + iy$ satisfies $|z|-2=0$ and $|z-i|-|z+5 i|=0$, then
→The eccentricity of a hyperbola passing through the points $(3, 0)$, $(3\sqrt 2 ,\;2)$ will be
→If $\alpha>\beta>\gamma>0$, then the expression
$\cot ^{-1}\left\{\beta+\frac{\left(1+\beta^{2}\right)}{(\alpha-\beta)}\right\}+\cot ^{-1}\left\{\gamma+\frac{\left(1+\gamma^{2}\right)}{(\beta-\gamma)}\right\}$ $+\cot ^{-1}\left\{\alpha+\frac{\left(1+\alpha^{2}\right)}{(\gamma-\alpha)}\right\}$ is equal to:
→$\cot ^{-1}\left\{\beta+\frac{\left(1+\beta^{2}\right)}{(\alpha-\beta)}\right\}+\cot ^{-1}\left\{\gamma+\frac{\left(1+\gamma^{2}\right)}{(\beta-\gamma)}\right\}$ $+\cot ^{-1}\left\{\alpha+\frac{\left(1+\alpha^{2}\right)}{(\gamma-\alpha)}\right\}$ is equal to: