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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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}

Vedclass · Accepted Answer

$\frac{\pi}{4}$

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} = $

Similar Questions

$\int_0^\pi x \sin x \cos^4 x \, dx = $

$\int_0^{\pi /2} \log(\sin x) \, dx = $

If $I_{n} = \int_{0}^{\frac{\pi}{4}} \tan^{n} x \, dx$,where $n$ is a positive integer,then $I_{10} + I_{8}$ is equal to

If $f(x) = \frac{x^3+5}{\sqrt{12+x}}$ and $\int_{-5}^5 f(x) dx = \int_0^5 (f(x) + g(x)) dx$,then $g(x) =$

Value of the definite integral,$\int\limits_{ - \frac{1}{2}}^{\frac{1}{2}} {\,(\,\,{{\sin }^{ - 1}}(3x - 4{x^3})\,\, - \,\,{{\cos }^{ - 1}}(4{x^3} - 3x)\,\,)\,dx} \,\,$

Vedclass Test Series

Exam Paper Generator

Online Exam Module