If int_{-pi / 2}^{pi / 2} frac{8 sqrt{2} cos | vedclass

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

Question

(D) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{8 \sqrt{2} \cos x}{(1+e^{\sin x})(1+\sin ^4 x)} \, dx$ $(1)$Using the property $\int_a^b f(x)

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(D) Let $I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{8 \sqrt{2} \cos x}{(1+e^{\sin x})(1+\sin ^4 x)} \, dx$ $(1)$Using the property $\int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx$,we get:$I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{8 \sqrt{2} \cos x}{(1+e^{-\sin x})(1+\sin ^4 x)} \, dx = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{8 \sqrt{2} \cos x \cdot e^{\sin x}}{(1+e^{\sin x})(1+\sin ^4 x)} \, dx$ $(2)$Adding $(1)$ and $(2)$:$2I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{8 \sqrt{2} \cos x}{1+\sin ^4 x} \, dx = 2 \int_{0}^{\frac{\pi}{2}} \frac{8 \sqrt{2} \cos x}{1+\sin ^4 x} \, dx$$I = \int_{0}^{\frac{\pi}{2}} \frac{8 \sqrt{2} \cos x}{1+\sin ^4 x} \, dx$Let $\sin x = t$,then $\cos x \, dx = dt$. Limits change from $0$ to $1$:$I = \int_{0}^{1} \frac{8 \sqrt{2}}{1+t^4} \, dt = 4 \sqrt{2} \int_{0}^{1} \frac{2}{1+t^4} \, dt = 4 \sqrt{2} \int_{0}^{1} \frac{(t^2+1) - (t^2-1)}{1+t^4} \, dt$$I = 4 \sqrt{2} \left[ \int_{0}^{1} \frac{1+1/t^2}{t^2+1/t^2} \, dt - \int_{0}^{1} \frac{1-1/t^2}{t^2+1/t^2} \, dt ight]$$I = 4 \sqrt{2} \left[ \int_{0}^{1} \frac{d(t-1/t)}{(t-1/t)^2+2} - \int_{0}^{1} \frac{d(t+1/t)}{(t+1/t)^2-2} ight]$Evaluating the integrals:$I = 4 \sqrt{2} \left[ \frac{1}{\sqrt{2}} 	an^{-1} \left( \frac{t-1/t}{\sqrt{2}} ight) \Big|_0^1 - \frac{1}{2\sqrt{2}} \ln \left| \frac{t+1/t-\sqrt{2}}{t+1/t+\sqrt{2}} ight| \Big|_0^1 ight]$$I = 4 \sqrt{2} \left[ \frac{1}{\sqrt{2}} (\frac{\pi}{2} - (-\frac{\pi}{2})) - \frac{1}{2\sqrt{2}} (\ln \frac{2-\sqrt{2}}{2+\sqrt{2}} - \ln 1) ight]$$I = 2\pi - 2 \ln \left( \frac{2-\sqrt{2}}{2+\sqrt{2}} \cdot \frac{2-\sqrt{2}}{2-\sqrt{2}} ight) = 2\pi - 2 \ln \left( \frac{6-4\sqrt{2}}{2} ight) = 2\pi - 2 \ln (3-2\sqrt{2}) = 2\pi + 2 \ln (3+2\sqrt{2})$Thus,$\alpha = 2, \beta = 2$. Therefore,$\alpha^2 + \beta^2 = 2^2 + 2^2 = 8$.

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.....................

Similar Questions

The value of $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}(x^3+\cos x+\tan^5 x) dx$ is equal to . . . . . . .

If $I = \int_0^{\frac{\pi}{4}} \log (1 + \tan x) \, dx$,then the value of $I$ is

The value of $\int_2^3 {\frac{{\sqrt x }}{{\sqrt {5 - x} + \sqrt x }}} \,dx$ is

If $\int_{0}^{100 \pi} \frac{\sin ^{2} x}{e^{\left(\frac{x}{\pi}-\left[\frac{x}{\pi}\right]\right)}} d x=\frac{\alpha \pi^{3}}{1+4 \pi^{2}}, \alpha \in R$,where $[x]$ is the greatest integer less than or equal to $x$,then the value of $\alpha$ is :

The value of $\int_0^{n\pi + v} {|\sin x|\,dx} $ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module