নির্দিষ্ট যোগজ
∫0πtanxsecx+cosxdx=\displaystyle \int_{0}^{\pi}\frac{\tan x}{\sec x+\cos x}dx_{=}∫0πsecx+cosxtanxdx=
π\piπ
−π2\dfrac {-\pi}22−π
−π-\pi−π
2π2 \pi2π
∫0πtanxsecx+cosxdx\int_{0}^{\pi}\dfrac{\tan x}{\sec x+\cos x}dx∫0πsecx+cosxtanxdx
∫tanxsecx+cosxdx=∫d(secx)1+sec2x\int \dfrac{\tan x}{\sec x+\cos x}dx=\int \dfrac{d (\sec x)}{1+\sec ^2 x}∫secx+cosxtanxdx=∫1+sec2xd(secx)
=tan−1(secx)+c=\tan ^{-1}(\sec x)+c=tan−1(secx)+c
∫0πtanxsecx+cosxdx=[tan(secx)+c]∣0π\int_{0}^{\pi}\dfrac{\tan x}{\sec x+\cos x}dx=\left [ \tan (\sec x)+c \right ] |_{0}^{\pi}∫0πsecx+cosxtanxdx=[tan(secx)+c]∣0π
=(tan−1(secπ)+c)−[tan−1(sec0)]=(\tan ^{-1}(\sec \pi)+c)- [ \tan ^{-1}(\sec 0) ]=(tan−1(secπ)+c)−[tan−1(sec0)]
=−π2=-\dfrac{\pi}{2}=−2π
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?
∫0π6sin2xcosxdx= \int_{0}^{\frac{\pi}{6}} \sin ^{2} x \cos x d x= ∫06πsin2xcosxdx= কত ?
f(x)= {x+1forx=0 \left \lbrace \begin{matrix} x + 1 & f{\quad\text{or}\quad} & x & = & 0 \end{matrix} \right . {x+1forx=0 হলে-
∫−1−12f(x)dx=18 \int_{- 1}^{- \frac{1}{2}} f{\left ( x \right )} dx = \frac{1}{8} ∫−1−21f(x)dx=81
∫01f(x)dx=0 \int_{0}^{1} f{\left ( x \right )} dx = 0 ∫01f(x)dx=0
f(−1)=1 f{\left ( - 1 \right )} = 1 f(−1)=1
নিচের কোনটি সঠিক?
The value of ;∫0π/4secx(secx+tanx)2dx\displaystyle \int_{0}^{\pi /4}\frac{\sec x}{\left ( \sec x+\tan x \right )^{2}}dx∫0π/4(secx+tanx)2secxdx is& ;
