ত্রিকোনোমিতিক ফাংশনের যোগজীকরণ
যোজিত ফল নির্ণয় করঃ
(a) ∫exx2+1(x+1)2dx \int e^{x} \frac{x^{2}+1}{(x+1)^{2}} d x ∫ex(x+1)2x2+1dx
(b) ∫sin5xdx \int \sin ^{5} x d x ∫sin5xdx
(a)∫exx2+1(x+1)2dx=∫exx2−1+2(x+1)2dx=∫ex{x−1x+1+2(x+1)2}dx=∫ex{x−1x+1+ddx(x−1x+1)}dx=[∫ex{f(x)+f′(x)}dx=exf(x)+c]=ex(x−1)x+1+c [Ans.] \begin{array}{l} \text{(a)} \int \mathrm{e}^{\mathrm{x}} \frac{\mathrm{x}^{2}+1}{(\mathrm{x}+1)^{2}} \mathrm{dx}=\int \mathrm{e}^{\mathrm{x}} \frac{\mathrm{x}^{2}-1+2}{(\mathrm{x}+1)^{2}} \mathrm{dx}=\int \mathrm{e}^{\mathrm{x}}\left\{\frac{\mathrm{x}-1}{\mathrm{x}+1}+\frac{2}{(\mathrm{x}+1)^{2}}\right\} \mathrm{dx} \\ =\int \mathrm{e}^{\mathrm{x}}\left\{\frac{\mathrm{x}-1}{\mathrm{x}+1}+\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{x}-1}{\mathrm{x}+1}\right)\right\} \mathrm{dx}=\left[\int \mathrm{e}^{\mathrm{x}}\left\{\mathrm{f}(\mathrm{x})+\mathrm{f}^{\prime}(\mathrm{x})\right\} \mathrm{dx}=\mathrm{e}^{\mathrm{x}} \mathrm{f}(\mathrm{x})+\mathrm{c}\right]=\mathrm{e}^{\mathrm{x}} \frac{(\mathrm{x}-1)}{\mathrm{x}+1}+\mathrm{c} \text { [Ans.] }\end{array} (a)∫ex(x+1)2x2+1dx=∫ex(x+1)2x2−1+2dx=∫ex{x+1x−1+(x+1)22}dx=∫ex{x+1x−1+dxd(x+1x−1)}dx=[∫ex{f(x)+f′(x)}dx=exf(x)+c]=exx+1(x−1)+c [Ans.]
(b)∫sin5xdx=∫(sin2x)2sinxdx=∫(1−cos2x)2sinxdx=∫(1−z2)2(−dz)[cosx=z रनে, dz=−sinxdx]=−∫(1−2z2+z4)dz=∫(−1+2z2−z4)dz=−z+23z3−15z5+c=−cosx+23cos3x−15cos5x+c [Ans.] \begin{array}{l}\text{(b)} \int \sin ^{5} \mathrm{xdx}=\int\left(\sin ^{2} \mathrm{x}\right)^{2} \sin \mathrm{xdx}=\int\left(1-\cos ^{2} \mathrm{x}\right)^{2} \sin \mathrm{xdx} \\ =\int\left(1-\mathrm{z}^{2}\right)^{2}(-\mathrm{dz}) \quad[\cos \mathrm{x}=\mathrm{z} \text { रनে, } \mathrm{dz}=-\sin \mathrm{xdx}] \\ =-\int\left(1-2 \mathrm{z}^{2}+\mathrm{z}^{4}\right) \mathrm{dz}=\int\left(-1+2 \mathrm{z}^{2}-\mathrm{z}^{4}\right) \mathrm{dz}=-\mathrm{z}+\frac{2}{3} \mathrm{z}^{3}-\frac{1}{5} \mathrm{z}^{5}+\mathrm{c} \\ =-\cos \mathrm{x}+\frac{2}{3} \cos ^{3} \mathrm{x}-\frac{1}{5} \cos ^{5} \mathrm{x}+\mathrm{c} \quad \text { [Ans.] }\end{array} (b)∫sin5xdx=∫(sin2x)2sinxdx=∫(1−cos2x)2sinxdx=∫(1−z2)2(−dz)[cosx=z रनে, dz=−sinxdx]=−∫(1−2z2+z4)dz=∫(−1+2z2−z4)dz=−z+32z3−51z5+c=−cosx+32cos3x−51cos5x+c [Ans.]
f(x)=x………(i) f(x)=x \ldots \ldots \ldots(i) f(x)=x………(i)
g(x)=cos−1x2………(ii) g(x)=\cos ^{-1} x^2 \ldots \ldots \ldots(i i) g(x)=cos−1x2………(ii)
y2=7x………(iii) y^2=7 x \ldots \ldots \ldots(i i i) y2=7x………(iii)
∫dx1+cosx=f(x)+c \int \frac{dx}{1 + \cos{x}} = f{\left ( x \right )} + c ∫1+cosxdx=f(x)+c হলে, f(x)=?
∫1−cos4xdx=?\int \sqrt{1-\cos 4 x} d x = ?∫1−cos4xdx=?
∫cos−1xdx \int \cos^{- 1}{x} dx ∫cos−1xdx এর মান কোনটি?
