Question
${x^5} + 10{x^4}a + 40{x^3}{a^2} + 80{x^2}{a^3}$$ + 80x{a^4} + 32{a^5} = $
अत:${(x + 2a)^5} = {x^5} + 10{x^4}a + 40{x^3}{a^2} + 80{x^2}{a^3} + 80x{a^4} + 32{a^5}$.
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$\int \frac{\sin \theta \cdot \sin 2 \theta\left(\sin ^{6} \theta+\sin ^{4} \theta+\sin ^{2} \theta\right) \sqrt{2 \sin ^{4} \theta+3 \sin ^{2} \theta+6}}{1-\cos 2 \theta} d \theta$ बराबर है
(जहाँ $c$ एक समाकलन अचर है)