int_{frac{pi}{6}}^{frac{pi}{3}} frac{1}{1+tan | vedclass

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

Question

(B) Let $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+	an^4 x} dx$. Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,where

Vedclass · Accepted Answer

$\frac{\pi}{12}$

Vedclass · Answer

(B) Let $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+	an^4 x} dx$. Using the property $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x) dx$,where $a+b = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}$,we have: $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+	an^4(\frac{\pi}{2}-x)} dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+\cot^4 x} dx$. $I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{	an^4 x}{1+	an^4 x} dx$. Adding the two expressions for $I$: $2I = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1+	an^4 x}{1+	an^4 x} dx = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1 dx$. $2I = [x]_{\frac{\pi}{6}}^{\frac{\pi}{3}} = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$. Therefore,$I = \frac{\pi}{12}$.

$\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+\tan^4 x} dx = $ . . . . . . .

Similar Questions

$\int_{0}^{1} x(1 - x)^{5} dx = . . . . . .$

$\int_0^{\frac{\pi}{2}} \frac{\sin^2 x}{\sin x + \cos x} dx =$

Let $f(x) = \frac{\sin x}{x}$,then $\int_{0}^{\frac{\pi}{2}} f(x) f\left(\frac{\pi}{2} - x\right) dx =$

By using the properties of definite integrals,evaluate the integral $\int_{0}^{2} x \sqrt{2-x} \, dx$.

$\int\limits_{\frac{1}{2}}^{3\frac{1}{2}} {\left\{ {\frac{1}{2}\,\left( {|x - 3| + |1 - x| - 4} \right)} \right\}\,dx} $ equals: Where $\{*\}$ denotes the fractional part function.

Vedclass Test Series

Exam Paper Generator

Online Exam Module