UV আকারের (Integration by parts)
The integral of ∫esinx(xcosx−secxtanx)dx\displaystyle\int e^{\sin x}(x\cos x-\sec x\tan x)dx∫esinx(xcosx−secxtanx)dx is?
xesinx−esinxsecx+cxe^{\sin x}-e^{\sin x}\sec x+cxesinx−esinxsecx+c
(x+secx)esinx+c(x+\sec x)e^{\sin x}+c(x+secx)esinx+c
esinxcosx+ce^{\sin x}\cos x+cesinxcosx+c
esinx(cosx−secx)+ce^{\sin x}(\cos x-\sec x)+cesinx(cosx−secx)+c
∫esinx⋅x⋅cosxdx−∫esinx⋅secx⋅tanx⋅dx=∫xf(x)esinx⋅cosxg(x)⋅dx−∫ef(x)sinx⋅secx⋅tanx⋅dx⏟g(x) \begin{aligned} & \int e^{\sin x \cdot x \cdot \cos x} d x-\int e^{\sin x} \cdot \sec x \cdot \tan x \cdot d x \\ = & \int \frac{x}{f(x)} e^{\sin x} \cdot \frac{\cos x}{g(x)} \cdot d x-\int e_{f(x)}^{\sin x} \cdot \underbrace{\sec x \cdot \tan x \cdot d x}_{g(x)}\end{aligned} =∫esinx⋅x⋅cosxdx−∫esinx⋅secx⋅tanx⋅dx∫f(x)xesinx⋅g(x)cosx⋅dx−∫ef(x)sinx⋅g(x)secx⋅tanx⋅dx
Using ILATE.
=x⋅esinx−∫1⋅esinxdx−∫secx⋅tanx⋅esinx+∫esinx⋅cosx⋅sesx⋅dx=x⋅esinx−∫esinx⋅dx−esinx⋅secx+∫esinx⋅dx=x⋅esinx−esinxsecx+c \begin{array}{l} =x \cdot e^{\sin x}-\int 1 \cdot e^{\sin x} d x-\int \sec x \cdot \tan x \cdot e^{\sin x} \\ \quad+\int e^{\sin x} \cdot \cos x \cdot \operatorname{ses} x \cdot d x \\ =x \cdot e^{\sin x}-\int e^{\sin x} \cdot d x-e^{\sin x} \cdot \sec x+\int e^{\sin x} \cdot d x \\ =x \cdot e^{\sin x}-e^{\sin x} \sec x+c \end{array} =x⋅esinx−∫1⋅esinxdx−∫secx⋅tanx⋅esinx+∫esinx⋅cosx⋅sesx⋅dx=x⋅esinx−∫esinx⋅dx−esinx⋅secx+∫esinx⋅dx=x⋅esinx−esinxsecx+c
∫x3exdx=f(x)+c \int x^{3} e^{x} dx = f{\left ( x \right )} + c ∫x3exdx=f(x)+c হয় তবে f(x)=?
Evaluate: ∫xxln(ex)dx\int x ^ { x } \ln ( e x ) d x∫xxln(ex)dx
∫f(x)dx=x+e2x \int f{\left ( x \right )} dx = x + e^{2 x} ∫f(x)dx=x+e2x হলে f(x)=?
The integrating factor of the differential equation dydx(xloge x)+y=2loge x\dfrac{dy}{dx}\left(x\log _e\:x\right)+y=2\log _e\:xdxdy(xlogex)+y=2logex is given by
