প্রতিস্থাপন পদ্ধতি (Method of Substitution)
∫tan(sin−1x)1−x2dx=?\int_{ }^{ }\frac{\tan\left(\sin^{-1}x\right)}{\sqrt{1-x^2}}dx=?∫1−x2tan(sin−1x)dx=?
sec2(sin−1x)+c\sec^2(\sin^{-1}x)+csec2(sin−1x)+c
sec(sin−1x)+c\sec(\sin^{-1}x)+csec(sin−1x)+c
ln∣sec(sin−1x)∣+c\ln|\sec(\sin^{-1}x)|+cln∣sec(sin−1x)∣+c
ln∣tan(sin−1x)∣+c\ln|\tan(\sin^{-1}x)|+cln∣tan(sin−1x)∣+c
∫tan(sin−1x)1−x2dx\int_{ }^{ }\frac{\tan\left(\sin^{-1}x\right)}{\sqrt{1-x^2}}dx∫1−x2tan(sin−1x)dx
=∫tan(sin−1x).11−x2dx=∫tan(sin−1x).d(sin−1x)=\int_{ }^{ }\tan\left(\sin^{-1}x\right).\frac{1}{\sqrt{1-x^2}}dx=\int_{ }^{ }\tan(\sin^{-1}x).d(\sin^{-1}x)=∫tan(sin−1x).1−x21dx=∫tan(sin−1x).d(sin−1x)
=ln∣sec(sin−1x)∣+c=\ln|\sec(\sin^{-1}x)|+c=ln∣sec(sin−1x)∣+c
∫ex(x+1)dxsin2(xex)=? \int \frac{e^{x} \left ( x + 1 \right ) dx}{\sin^{2}{\left ( x e^{x} \right )}} = ? ∫sin2(xex)ex(x+1)dx=?
∫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?
∫sinx3+4cosxdx=?\int \frac{\sin x}{3+4 \cos x} d x = ?∫3+4cosxsinxdx=?
