MHB Harish Karakoti's question on Facebook

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Harish Karakoti inquired about evaluating the integral I = ∫ from -π to π of (x^4 cos(x)) / (1 - sin(x) + √(1 + sin²(x))) dx. The integral was separated into two intervals, allowing for a change of variable in the second integral. By combining the two resulting integrals and simplifying, it was shown that the sum equals ∫ from 0 to π of x^4 cos(x) dx. Using integration by parts, the final result was determined to be I = 24π - 4π³. This method effectively demonstrates the evaluation of the complex integral.
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Harish Karakoti asked how to find the integral

$$I = \int^{\pi}_{-\pi} \frac{x^4\cos(x)}{1-\sin(x)+\sqrt{1+\sin^2(x)}}dx$$
 
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$$I = \int^{\pi}_{-\pi} \frac{x^4\cos(x)}{1-\sin(x)+\sqrt{1+\sin^2(x)}}dx $$

Separate the intervals

$$= \int^{\pi}_{0} \frac{x^4\cos(x)}{1-\sin(x)+\sqrt{1+\sin^2(x)}}dx-\int^{-\pi}_{-0} \frac{x^4\cos(x)}{1-\sin(x)+\sqrt{1+\sin^2(x)}}dx$$

use change of variable in the second one
$$I= \int^{\pi}_{0} \frac{x^4\cos(x)}{1-\sin(x)+\sqrt{1+\sin^2(x)}}dx+\int^{\pi}_{0} \frac{x^4\cos(x)}{1+\sin(x)+\sqrt{1+\sin^2(x)}}dx $$

By combining the two integrals, and noticing that

$$\frac{1}{1-\sin(x)+\sqrt{1+\sin^2(x)}}+ \frac{1}{1+\sin(x)+\sqrt{1+\sin^2(x)}} =1$$

We get

$$I=\int^{\pi}_{0} x^4\cos(x)\,dx $$

Use integration by parts to conclude that
$$I=\int^{\pi}_{0} x^4\cos(x)\,dx =24\pi -4\pi^3 $$
 
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