নির্দিষ্ট যোগজ
∫0∞x2015(1+x)2017dx=?\int_{0}^{\infty}{\frac{x^{2015}}{\left(1+x\right)^{2017}}dx=?} ∫0∞(1+x)2017x2015dx=?
12008\frac{1}{2008} 20081
12017\frac{1}{2017} 20171
2
12016\frac{1}{2016} 20161
∫0∞x2015(1+x)2017dx\int_{0}^{\infty}{\frac{x^{2015}}{\left(1+x\right)^{2017}}dx} ∫0∞(1+x)2017x2015dx
=−∫0∞− 1x2dx(1+1x)2017=[(1+1x)−20162016]0∞=12016-\int_{0}^{\infty}\frac{-\ \ \frac{1}{x^2}dx}{\left(1+\frac{1}{x}\right)^{2017}}=\left[\frac{\left(1+\frac{1}{x}\right)^{-2016}}{2016}\right]_0^\infty=\frac{1}{2016} −∫0∞(1+x1)2017− x21dx=[2016(1+x1)−2016]0∞=20161
∫0π4cosθcos2θdθ=? \int_{0}^{\frac{\pi}{4}} \frac{\cos \theta}{\cos ^{2} \theta} d \theta=? ∫04πcos2θcosθdθ=?
∫0π2cos3xsinxdx \int_{0}^{\frac{\pi}{2}} \cos ^{3} x \sqrt{\sin x} d x ∫02πcos3xsinxdx
The value of ∫−π/2199π/2(1+cos2x)dx\displaystyle\int^{199\pi/2}_{-\pi/2}\sqrt{(1+\cos 2x)}dx∫−π/2199π/2(1+cos2x)dx is?
∫oπ2dx1+cosx=? \int_{o}^{\frac{\pi}{2}} \frac{dx}{1 + \cos{x}} = ? ∫o2π1+cosxdx=?
