নির্দিষ্ট যোগজ
∫012x1+x2dx\int_0^1\frac{2x}{1+x^2}dx∫011+x22xdx এর মান-
0
1
ln3\ln3ln3
ln2\ln2ln2
∫012x1+x2dx \int_{0}^{1} \frac{2 x}{1+x^{2}} d x ∫011+x22xdx
Let, 1+x2=t 1+x^{2}=t 1+x2=t
or 2xdx=dt 2 x d x=d t 2xdx=dt
Put x=0 x=0 x=0 or t=1 t=1 t=1
and x=1 x=1 x=1 or t=2 t=2 t=2
or I=∫12dtt=[loget]12 I=\int_{1}^{2} \frac{d t}{t}=\left[\log _{e} t\right]_{1}^{2} I=∫12tdt=[loget]12
=loge2−loge1=loge(21)=loge2 \begin{array}{l} =\log _{e} 2-\log _{e} 1 \\ =\log _{e}\left(\frac{2}{1}\right)=\log _{e} 2 \end{array} =loge2−loge1=loge(12)=loge2
∫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?
∫0π4cos2θcos2θ=?\int_0^{\frac{\pi}{4}} \frac{cos 2 \theta}{cos² \theta} = ? ∫04πcos2θcos2θ=?
