Replacement সংক্রান্ত
যোজিত ফল নির্ণয় কর: ∫adx((x2+a2)3 \int \frac{a dx}{\left ( \left ( \sqrt{x^{2} + a^{2}} \right )^{3} \right.} ∫((x2+a2)3adx
∫adx(x2+a2)3; let, x=atanθ∴dx=asec2θdθ=∫a2sec2θdθ{a2(1+tan2θ)}3/2=∫a2sec2θdθa3sec3θ=1a∫cosθdθ=1asinθ+c=1asin[tan−1(xa)]+c [Ans.] \begin{array}{l} \int \frac{\mathrm{adx}}{\left(\sqrt{\mathrm{x}^{2}+\mathrm{a}^{2}}\right)^{3}} ; \text { let, } \mathrm{x}=\mathrm{a} \tan \theta \therefore \mathrm{dx}=\mathrm{a} \mathrm{sec}^{2} \theta \mathrm{d} \theta \\ =\int \frac{\mathrm{a}^{2} \sec ^{2} \theta \mathrm{d} \theta}{\left\{\mathrm{a}^{2}\left(1+\tan ^{2} \theta\right)\right\}^{3 / 2}}=\int \frac{\mathrm{a}^{2} \sec ^{2} \theta \mathrm{d} \theta}{\mathrm{a}^{3} \sec ^{3} \theta}=\frac{1}{\mathrm{a}} \int \cos \theta \mathrm{d} \theta=\frac{1}{\mathrm{a}} \sin \theta+\mathrm{c}=\frac{1}{\mathrm{a}} \sin \left[\tan ^{-1}\left(\frac{\mathrm{x}}{\mathrm{a}}\right)\right]+\mathrm{c} \text { [Ans.] }\end{array} ∫(x2+a2)3adx; let, x=atanθ∴dx=asec2θdθ=∫{a2(1+tan2θ)}3/2a2sec2θdθ=∫a3sec3θa2sec2θdθ=a1∫cosθdθ=a1sinθ+c=a1sin[tan−1(ax)]+c [Ans.]
∫x1+x2dx=f(x) \int \frac{x}{\sqrt{1 + x^{2}}} dx = f{\left ( x \right )} ∫1+x2xdx=f(x) হলে, f(x)=কত?
∫72cos8x+3x2x3+9sin8xdx= \int \frac{72 \cos 8 x+3 x^{2}}{x^{3}+9 \sin 8 x} d x= ∫x3+9sin8x72cos8x+3x2dx= কত ?
∫1x2sin(1x)dx \int \frac{1}{x^{2}} \sin \left(\frac{1}{x}\right) d x ∫x21sin(x1)dx এর মান কত?