UV আকারের (Integration by parts)
∫sin4xsin2xdx=?\int \sin 4 x \sin 2 x d x \quad = ?∫sin4xsin2xdx=?
14cos2x−112cos6x+c\frac{1}{4} \cos 2 x-\frac{1}{12} \cos 6 x+c41cos2x−121cos6x+c
14sin2x−112sin6x+c\frac{1}{4} \sin 2 x-\frac{1}{12} \sin 6 x+c41sin2x−121sin6x+c
14cos2x−112sin6x+c\frac{1}{4} \cos 2 x-\frac{1}{12} \sin 6 x+c41cos2x−121sin6x+c
14sin2x−112cos6x+c\frac{1}{4} \sin 2 x-\frac{1}{12} \cos 6 x+c41sin2x−121cos6x+c
Solve:
∫sin4xsin2xdx=∫12{cos(4x−2x)−cos(4x+2x)}dx=12∫(cos2x−cos6x)dx=12(sin2x2−sin6x6)+c=14sin2x−112sin6x+c \begin{array}{l} \int \sin 4 x \sin 2 x d x \quad \\ =\int \frac{1}{2}\{\cos (4 x-2 x)-\cos (4 x+2 x)\} d x \\ =\frac{1}{2} \int(\cos 2 x-\cos 6 x) d x \\ =\frac{1}{2}\left(\frac{\sin 2 x}{2}-\frac{\sin 6 x}{6}\right)+c \\ =\frac{1}{4} \sin 2 x-\frac{1}{12} \sin 6 x+c \end{array} ∫sin4xsin2xdx=∫21{cos(4x−2x)−cos(4x+2x)}dx=21∫(cos2x−cos6x)dx=21(2sin2x−6sin6x)+c=41sin2x−121sin6x+c
∫e2xcosexdx=?\int e^{2 x} \cos e^{x} d x = ?∫e2xcosexdx=?
∫xcos−1x2dx=? \int x \cos ^{-1} x^{2} d x = ?∫xcos−1x2dx=?
∫x3sinxdx=?\int x^{3} \sin x d x = ?∫x3sinxdx=?
∫ln(1+x)1+xdx equals\displaystyle\int {\dfrac{{\ln \left( {1 + {x}} \right)}}{{1 + {x}}}} dx\,equals∫1+xln(1+x)dxequals
