int_0^{pi /2} frac{dtheta}{1 + tan theta} = | vedclass

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(D) Let $I = \int_0^{\pi /2} \frac{d	heta}{1 + 	an 	heta}$.Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we have:$I = \int_0^{\pi /2}

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$\frac{\pi}{4}$

Vedclass · Answer

(D) Let $I = \int_0^{\pi /2} \frac{d	heta}{1 + 	an 	heta}$.Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we have:$I = \int_0^{\pi /2} \frac{d	heta}{1 + 	an(\frac{\pi}{2} - 	heta)} = \int_0^{\pi /2} \frac{d	heta}{1 + \cot 	heta}$.Since $\cot 	heta = \frac{1}{	an 	heta}$,this becomes:$I = \int_0^{\pi /2} \frac{d	heta}{1 + \frac{1}{	an 	heta}} = \int_0^{\pi /2} \frac{	an 	heta}{1 + 	an 	heta} d	heta$.Adding the two expressions for $I$:$2I = \int_0^{\pi /2} \left( \frac{1}{1 + 	an 	heta} + \frac{	an 	heta}{1 + 	an 	heta} ight) d	heta$.$2I = \int_0^{\pi /2} \frac{1 + 	an 	heta}{1 + 	an 	heta} d	heta = \int_0^{\pi /2} 1 d	heta$.$2I = [	heta]_0^{\pi /2} = \frac{\pi}{2}$.Therefore,$I = \frac{\pi}{4}$.

$\int_0^{\pi /2} \frac{d\theta}{1 + \tan \theta} = $

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