নির্দিষ্ট যোগজ
∫0π4cosθcos2θdθ=? \int_{0}^{\frac{\pi}{4}} \frac{\cos \theta}{\cos ^{2} \theta} d \theta=? ∫04πcos2θcosθdθ=?
ln(1+√2)
ln(1-√2)
I=∫0π4cosθcos2θdθ=∫0π41cosθdθ=∫0π4secθdθ=[ln∣tanθ+secθ∣]0π4=ln∣tanπ4+secπ4∣−ℓn∣tan0+sec0∣=ℓn∣1+2∣−ℓn1=ℓn∣1+2∣−0 \begin{array}{l} \mathrm{I}=\int_{0}^{\frac{\pi}{4}} \frac{\cos \theta}{\cos ^{2} \theta} \mathrm{d} \theta=\int_{0}^{\frac{\pi}{4}} \frac{1}{\cos \theta} \mathrm{d} \theta=\int_{0}^{\frac{\pi}{4}} \sec \theta \mathrm{d} \theta=[\ln |\tan \theta+\sec \theta|]_{0}^{\frac{\pi}{4}} \\ =\ln \left|\tan \frac{\pi}{4}+\sec \frac{\pi}{4}\right|-\ell \mathrm{n}|\tan 0+\sec 0|=\ell \mathrm{n}|1+\sqrt{2}|-\ell \mathrm{n} 1=\ell \mathrm{n}|1+\sqrt{2}|-0\end{array} I=∫04πcos2θcosθdθ=∫04πcosθ1dθ=∫04πsecθdθ=[ln∣tanθ+secθ∣]04π=lntan4π+sec4π−ℓn∣tan0+sec0∣=ℓn∣1+2∣−ℓn1=ℓn∣1+2∣−0
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?
∫0π/2cosxdx= কত? \int_{0}^{\pi / 2} \cos x d x=\text { কত? } ∫0π/2cosxdx= কত?
∫0π6sin2xcosxdx= \int_{0}^{\frac{\pi}{6}} \sin ^{2} x \cos x d x= ∫06πsin2xcosxdx= কত ?
∫01e−(x2)dx=?\int_{0}^{1} e^{- \left ( x^{2} \right )} dx = ?∫01e−(x2)dx=?