প্রতিস্থাপন পদ্ধতি (Method of Substitution)
∫cosxdx(1−sinx)2=? \int \frac{\cos x d x}{(1-\sin x)^{2}} \quad = ?∫(1−sinx)2cosxdx=?
11−sinx+c\frac{1}{1-\sin x}+c1−sinx1+c
11−sin2x+c\frac{1}{1-\sin^2 x}+c1−sin2x1+c
11−cosx+c\frac{1}{1-\cos x}+c1−cosx1+c
11−cosx2+c\frac{1}{1-\cos x^2}+c1−cosx21+c
Solve: ধরি, 1−sinx=z 1-\sin x=z 1−sinx=z. তাহলে, −cosdx=dz -\cos d x=d z −cosdx=dz এবং
∫cosxdx(1−sinx)2=−∫dzz2=−∫z−2dz=−z−2+1−2+1+c=z−1+c=11−sinx+c \begin{array}{l} \int \frac{\cos x d x}{(1-\sin x)^{2}}=-\int \frac{d z}{z^{2}}=-\int z^{-2} d z \\ =-\frac{z^{-2+1}}{-2+1}+c=z^{-1}+c=\frac{1}{1-\sin x}+c \end{array} ∫(1−sinx)2cosxdx=−∫z2dz=−∫z−2dz=−−2+1z−2+1+c=z−1+c=1−sinx1+c
∫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?
যোগজীকরণ কর।
∫cosx3+cos2xdx \int \frac{\cos x}{3+\cos ^{2} x} d x ∫3+cos2xcosxdx
