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Exercise 9.10 · Q10

Q.The value of ∫0πdx1+5cos⁡x\displaystyle\int_0^\pi \dfrac{dx}{1+5^{\cos x}} is

(1) π2\dfrac{\pi}{2}
(2) π\pi
(3) 3π2\dfrac{3\pi}{2}
(4) 2π2\pi
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Applying the a+b−xa+b-x reflection with a=0,b=πa=0,b=\pi turns 5cos⁡x5^{\cos x} into 5−cos⁡x5^{-\cos x}; adding this reflected integral to the original produces a constant integrand that integrates trivially, then solving for II halves the result.

Step 1. Name the integral. Let I=∫0πdx1+5cos⁡xI=\displaystyle\int_0^\pi\frac{dx}{1+5^{\cos x}}.

Step 2. Reflect x→π−xx\to\pi-x (Property 6). Since cos⁡(π−x)=−cos⁡x\cos(\pi-x)=-\cos x,

I=∫0πdx1+5cos⁡(π−x)=∫0πdx1+5−cos⁡x.I=\int_0^\pi\frac{dx}{1+5^{\cos(\pi-x)}}=\int_0^\pi\frac{dx}{1+5^{-\cos x}}.

Step 3. Simplify the reflected integrand. Multiply numerator and denominator by 5cos⁡x5^{\cos x}:

11+5−cos⁡x=5cos⁡x5cos⁡x+1.\frac{1}{1+5^{-\cos x}}=\frac{5^{\cos x}}{5^{\cos x}+1}.

So I=∫0π5cos⁡x1+5cos⁡x dxI=\displaystyle\int_0^\pi\frac{5^{\cos x}}{1+5^{\cos x}}\,dx — a second expression for the same II. …

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