প্রতিস্থাপন পদ্ধতি (Method of Substitution)
∫2xsin−1x21−x4dx \int \frac{2 x \sin ^{-1} x^{2}}{\sqrt{1-x^{4}}} d x ∫1−x42xsin−1x2dx
12(sin−1x2)+c\frac{1}{2}\left(\sin ^{-1} x^{2}\right)+c21(sin−1x2)+c
12(sinx2)2+c\frac{1}{2}\left(\sin x^{2}\right)^{2}+c21(sinx2)2+c
12(sin−1x2)2+c\frac{1}{2}\left(\sin ^{-1} x^{2}\right)^{2}+c21(sin−1x2)2+c
12(sin−1x)2+c\frac{1}{2}\left(\sin ^{-1} x\right)^{2}+c21(sin−1x)2+c
Solve: ধরি, sin−1x2=z \sin ^{-1} x^{2}=z sin−1x2=z
∴11−(x2)2⋅2xdx=dz⇒2xdx1−x4=dz∴∫2xsin−1x21−x4dx=∫zdz \begin{array}{l} \therefore \quad \frac{1}{\sqrt{1-\left(x^{2}\right)^{2}}} \cdot 2 x d x=d z \Rightarrow \frac{2 x d x}{\sqrt{1-x^{4}}}=d z \\ \therefore \quad \int \frac{2 x \sin ^{-1} x^{2}}{\sqrt{1-x^{4}}} d x=\int z d z \end{array} ∴1−(x2)21⋅2xdx=dz⇒1−x42xdx=dz∴∫1−x42xsin−1x2dx=∫zdz
=z22+c=12(sin−1x2)2+c (Ans.) =\frac{z^{2}}{2}+c=\frac{1}{2}\left(\sin ^{-1} x^{2}\right)^{2}+c \text { (Ans.) } =2z2+c=21(sin−1x2)2+c (Ans.)
∫exdx1+e2x=f(x)+c \int \frac{e^{x} dx}{1 + e^{2 x}} = f{\left ( x \right )} + c ∫1+e2xexdx=f(x)+c
হলে, f(x)=?
What is ∫x4−1x2x4+x2+1dx\displaystyle \int \dfrac{x^4 - 1}{x^2 \sqrt{x^4 + x^2 + 1}} dx∫x2x4+x2+1x4−1dx equal to?
∫ecos−1x1−x2dx \int \frac{e^{\cos^{- 1}{x}}}{\sqrt{1 - x ²}} dx ∫1−x2ecos−1xdx এর মান কত?
∫cosxdx(1−sinx)2=? \int \frac{\cos x d x}{(1-\sin x)^{2}} \quad = ?∫(1−sinx)2cosxdx=?
