Associative and Distributive integrals MCQ Quiz in বাংলা - Objective Question with Answer for Associative and Distributive integrals - বিনামূল্যে ডাউনলোড করুন [PDF]

ধারণা:

${\displaystyle {\int }_0^{\pi /2}{\displaystyle \frac{\sqrt{{\sin }^8\mathrm{x}}}{\sqrt{{\sin }^8\mathrm{x}}+\sqrt{{\cos }^8\mathrm{x}}}}\mathrm{d}\mathrm{x}}$

গণনা:

ধরি, I  = ${\displaystyle {\int }_0^{\pi /2}{\displaystyle \frac{\sqrt{{\sin }^8(\pi /2-\mathrm{x})}}{\sqrt{{\sin }^8(\pi /2-\mathrm{x})}+\sqrt{{\cos }^8(\pi /2-\mathrm{x})}}}\mathrm{d}\mathrm{x}}$ ----(i)

⇒ I = ${\displaystyle {\int }_0^{\pi /2}{\displaystyle \frac{\sqrt{{\cos }^8\mathrm{x}}}{\sqrt{{\sin }^8\mathrm{x}}+\sqrt{{\cos }^8\mathrm{x}}}}\mathrm{d}\mathrm{x}}$

⇒ I = ${\displaystyle \frac{\pi }{4}}$ ----(ii)

(i) এবং (ii) যোগ করে পাই,

⇒ 2I = ${\displaystyle {\int }_0^{\pi /2}{\displaystyle \frac{\sqrt{{\cos }^8\mathrm{x}}+\sqrt{\mathrm{s}\mathrm{i}{\mathrm{n}}^8\mathrm{x}}}{\sqrt{{\cos }^8\mathrm{x}}+\sqrt{{\sin }^8\mathrm{x}}}}\mathrm{d}\mathrm{x}}$

⇒ 2I = ${\displaystyle {\int }_0^{\pi /2}\mathrm{d}\mathrm{x}}$

⇒ 2I = $[\mathrm{x}{]}_0^{\frac{\pi }{2}}$

⇒ I = ${\displaystyle \frac{\pi }{4}}$

ধারণা:

${\displaystyle {\int }_0^{\pi /2}{\displaystyle \frac{\sqrt{{\sin }^8\mathrm{x}}}{\sqrt{{\sin }^8\mathrm{x}}+\sqrt{{\cos }^8\mathrm{x}}}}\mathrm{d}\mathrm{x}}$

গণনা:

ধরি, I  = ${\displaystyle {\int }_0^{\pi /2}{\displaystyle \frac{\sqrt{{\sin }^8(\pi /2-\mathrm{x})}}{\sqrt{{\sin }^8(\pi /2-\mathrm{x})}+\sqrt{{\cos }^8(\pi /2-\mathrm{x})}}}\mathrm{d}\mathrm{x}}$ ----(i)

⇒ I = ${\displaystyle {\int }_0^{\pi /2}{\displaystyle \frac{\sqrt{{\cos }^8\mathrm{x}}}{\sqrt{{\sin }^8\mathrm{x}}+\sqrt{{\cos }^8\mathrm{x}}}}\mathrm{d}\mathrm{x}}$

⇒ I = ${\displaystyle \frac{\pi }{4}}$ ----(ii)

(i) এবং (ii) যোগ করে পাই,

⇒ 2I = ${\displaystyle {\int }_0^{\pi /2}{\displaystyle \frac{\sqrt{{\cos }^8\mathrm{x}}+\sqrt{\mathrm{s}\mathrm{i}{\mathrm{n}}^8\mathrm{x}}}{\sqrt{{\cos }^8\mathrm{x}}+\sqrt{{\sin }^8\mathrm{x}}}}\mathrm{d}\mathrm{x}}$

⇒ 2I = ${\displaystyle {\int }_0^{\pi /2}\mathrm{d}\mathrm{x}}$

⇒ 2I = $[\mathrm{x}{]}_0^{\frac{\pi }{2}}$

⇒ I = ${\displaystyle \frac{\pi }{4}}$

ধারণা:

${\displaystyle {\int }_0^{\pi /2}{\displaystyle \frac{\sqrt{{\sin }^8\mathrm{x}}}{\sqrt{{\sin }^8\mathrm{x}}+\sqrt{{\cos }^8\mathrm{x}}}}\mathrm{d}\mathrm{x}}$

গণনা:

ধরি, I  = ${\displaystyle {\int }_0^{\pi /2}{\displaystyle \frac{\sqrt{{\sin }^8(\pi /2-\mathrm{x})}}{\sqrt{{\sin }^8(\pi /2-\mathrm{x})}+\sqrt{{\cos }^8(\pi /2-\mathrm{x})}}}\mathrm{d}\mathrm{x}}$ ----(i)

⇒ I = ${\displaystyle {\int }_0^{\pi /2}{\displaystyle \frac{\sqrt{{\cos }^8\mathrm{x}}}{\sqrt{{\sin }^8\mathrm{x}}+\sqrt{{\cos }^8\mathrm{x}}}}\mathrm{d}\mathrm{x}}$

⇒ I = ${\displaystyle \frac{\pi }{4}}$ ----(ii)

(i) এবং (ii) যোগ করে পাই,

⇒ 2I = ${\displaystyle {\int }_0^{\pi /2}{\displaystyle \frac{\sqrt{{\cos }^8\mathrm{x}}+\sqrt{\mathrm{s}\mathrm{i}{\mathrm{n}}^8\mathrm{x}}}{\sqrt{{\cos }^8\mathrm{x}}+\sqrt{{\sin }^8\mathrm{x}}}}\mathrm{d}\mathrm{x}}$

⇒ 2I = ${\displaystyle {\int }_0^{\pi /2}\mathrm{d}\mathrm{x}}$

⇒ 2I = $[\mathrm{x}{]}_0^{\frac{\pi }{2}}$

⇒ I = ${\displaystyle \frac{\pi }{4}}$