প্রতিস্থাপন পদ্ধতি (Method of Substitution)
∫sinx3+4cosxdx=?\int \frac{\sin x}{3+4 \cos x} d x = ?∫3+4cosxsinxdx=?
−14ln∣3+4cosx∣+c-\frac{1}{4} \ln |3+4 \cos x|+c −41ln∣3+4cosx∣+c
−13ln∣3+4cosx∣+c-\frac{1}{3} \ln |3+4 \cos x|+c −31ln∣3+4cosx∣+c
−14ln∣3−4cosx∣+c-\frac{1}{4} \ln |3-4 \cos x|+c −41ln∣3−4cosx∣+c
14ln∣3+4cosx∣+c\frac{1}{4} \ln |3+4 \cos x|+c 41ln∣3+4cosx∣+c
Solve: ধরি, I=∫sinx3+4cosxdx \mathrm{I}=\int \frac{\sin x}{3+4 \cos x} d x I=∫3+4cosxsinxdx এবং 3+4cosx=z 3+4 \cos x=z 3+4cosx=z. তাহলে, −4sinxdx=dz -4 \sin x d x=d z −4sinxdx=dzএবং I=−14∫dzz=−14ln∣3+4cosx∣+c I=-\frac{1}{4} \int \frac{d z}{z}=-\frac{1}{4} \ln |3+4 \cos x|+c I=−41∫zdz=−41ln∣3+4cosx∣+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?
∫esin−1x⋅dx1−x2=?\int e^{\sin^ {-1}x} \cdot \frac{d x}{\sqrt{1-x^{2}}} = ?∫esin−1x⋅1−x2dx=?
