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$\int_0^{\frac{\pi}{2}} \sin ^6 \cdot x \cos ^4 \cdot x d x=$
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Verified Answer
The correct answer is:
$\frac{3\pi}{512}$
$$
I=\int_0^{\pi / 2} \sin ^6 x \cos ^4 x d x
$$
Using Walli's Formula,
$$
I=\frac{5 \times 3 \times 1 \times 3 \times 1}{10 \times 8 \times 6 \times 4 \times 2} \times \frac{\pi}{2}=\frac{3 \pi}{512}
$$
I=\int_0^{\pi / 2} \sin ^6 x \cos ^4 x d x
$$
Using Walli's Formula,
$$
I=\frac{5 \times 3 \times 1 \times 3 \times 1}{10 \times 8 \times 6 \times 4 \times 2} \times \frac{\pi}{2}=\frac{3 \pi}{512}
$$
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