int_{frac{-pi}{2}}^{frac{pi}{2}}(x^5-x^3 cos | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $f(x) = x^5 - x^3 \cos x + \sin^3 x - 3$. We observe that $f(-x) = (-x)^5 - (-x)^3 \cos(-x) + \sin^3(-x) - 3 = -x^5 + x^3 \cos x - \sin^3 x -

Vedclass · Accepted Answer

$-3\pi$

Vedclass · Answer

(C) Let $f(x) = x^5 - x^3 \cos x + \sin^3 x - 3$. We observe that $f(-x) = (-x)^5 - (-x)^3 \cos(-x) + \sin^3(-x) - 3 = -x^5 + x^3 \cos x - \sin^3 x - 3$. This does not immediately show $f(x)$ is even or odd,so we split the integral: $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^5 \, dx - \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} x^3 \cos x \, dx + \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^3 x \, dx - \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 3 \, dx$. Since $x^5$,$x^3 \cos x$,and $\sin^3 x$ are odd functions,their integrals over the symmetric interval $[-\frac{\pi}{2}, \frac{\pi}{2}]$ are $0$. Thus,$I = 0 - 0 + 0 - \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 3 \, dx$. $I = -3 	imes [x]_{-\frac{\pi}{2}}^{\frac{\pi}{2}} = -3 	imes (\frac{\pi}{2} - (-\frac{\pi}{2})) = -3 	imes \pi = -3\pi$.

$\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}(x^5-x^3 \cos x+\sin^3 x-3) \, dx = $ . . . . . .

Similar Questions

$\int_0^{400 \pi} \sqrt{1-\cos 2 x} \, dx =$ (in $\sqrt{2}$)

If $\int_{-\pi / 2}^{\pi / 2} \frac{8 \sqrt{2} \cos x \, dx}{(1+e^{\sin x})(1+\sin ^4 x)} = \alpha \pi + \beta \log _e(3+2 \sqrt{2})$,where $\alpha, \beta$ are integers,then $\alpha^2+\beta^2$ equals.....................

Given $\int_{0}^{\frac{\pi}{2}} \frac{dx}{1 + \sin x + \cos x} = \ln 2$,then the value of the definite integral $\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{1 + \sin x + \cos x} dx$ is equal to

$\int_0^{\pi / 4} \log (1+\tan x) d x=$

$\int_{ - \pi /2}^{\pi /2} {\frac{{\sin x}}{{1 + {{\cos }^2}x}}{e^{ - {{\cos }^2}x}}dx} $ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module