নির্দিষ্ট যোগজ
let, z=sinx \mathrm{z}=\sin \mathrm{x} z=sinx; dz=cosxdx] \mathrm{dz}=\cos \mathrm{x} \mathrm{dx}] dz=cosxdx]
∫0π2cos3xsinxdx=∫0π2(1−sin2x)sinxcosxdx=∫01(1−z2)zdz=[z12+112+1−z52+152+1]01=[23z32−27z72]01=821 \begin{array}{l} \int_{0}^{\frac{\pi}{2}} \cos ^{3} x \sqrt{\sin x} d x=\int_{0}^{\frac{\pi}{2}}\left(1-\sin ^{2} x\right) \sqrt{\sin x} \cos x d x=\int_{0}^{1}\left(1-z^{2}\right) \sqrt{z} d z \\ =\left[\frac{z^{\frac{1}{2}+1}}{\frac{1}{2}+1}-\frac{z^{\frac{5}{2}+1}}{\frac{5}{2}+1}\right]_{0}^{1}=\left[\frac{2}{3} z^{\frac{3}{2}}-\frac{2}{7} z^{\frac{7}{2}}\right]_{0}^{1}=\frac{8}{21} \end{array} ∫02πcos3xsinxdx=∫02π(1−sin2x)sinxcosxdx=∫01(1−z2)zdz=[21+1z21+1−25+1z25+1]01=[32z23−72z27]01=218
The value of ∫−π/2199π/2(1+cos2x)dx\displaystyle\int^{199\pi/2}_{-\pi/2}\sqrt{(1+\cos 2x)}dx∫−π/2199π/2(1+cos2x)dx is?
∫0π6sin2xcosxdx= \int_{0}^{\frac{\pi}{6}} \sin ^{2} x \cos x d x= ∫06πsin2xcosxdx= কত ?
∫1/21dxx4x2−1\int_{1 / 2}^{1} \frac{d x}{x \sqrt{4 x^{2}-1}}∫1/21x4x2−1dx
f(x)= {x+1forx=0 \left \lbrace \begin{matrix} x + 1 & f{\quad\text{or}\quad} & x & = & 0 \end{matrix} \right . {x+1forx=0 হলে-
∫−1−12f(x)dx=18 \int_{- 1}^{- \frac{1}{2}} f{\left ( x \right )} dx = \frac{1}{8} ∫−1−21f(x)dx=81
∫01f(x)dx=0 \int_{0}^{1} f{\left ( x \right )} dx = 0 ∫01f(x)dx=0
f(−1)=1 f{\left ( - 1 \right )} = 1 f(−1)=1
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