The value of is equal to: [2026]

Question

The value of $\int_{0}^{20 π} (\sin^{4} x + \cos^{4} x) d x$ is equal to: [2026]

Smart Achievers · Accepted Answer

(3)

Let

$I = \int_{0}^{20 π} (\sin^{4} x + \cos^{4} x) d x$

$= (\int_{0}^{20 π} {(\sin^{2} x + \cos^{2} x)}^{2} - 2 \sin^{2} x \cos^{2} x) d x$

$= \int_{0}^{20 π} (1 - \frac{1}{2} \sin^{2} (2 x) d x$

Period of $\sin^{2} (2 x) is π / 2$

$∴ I = 40 \int_{0}^{π / 2} (\sin^{4} x + \cos^{4} x) d x$

$= 40 \int_{0}^{π / 2} \sin^{4} x d x + 40 \int_{0}^{π / 2} \cos^{4} x d x$

Now, $\int_{0}^{π / 2} \sin^{4} x d x = \int_{0}^{π / 2} \cos^{4} x d x$

$∴ I = 40 \times 2 \int_{0}^{π / 2} \sin^{4} x d x$

$= 80 \int_{0}^{π / 2} \sin^{4} x d x$

Using Wallis' formula

$\int_{0}^{π / 2} \sin^{4} x d x = \frac{3 \cdot 1}{4 \cdot 2} \times \frac{π}{2} = \frac{3 π}{16}$

$∴ I = 80 \times (\frac{3 π}{16}) = 15 π$

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