Evaluate the definite integral int_{0}^{frac{ | vedclass

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

Question

(B) We need to evaluate the integral $I = \int_{0}^{\frac{2}{3}} \frac{d x}{4+9 x^{2}}$.First,rewrite the integrand as $\int \frac{d x}{(2)^{2}+(3 x)^

Vedclass · Accepted Answer

$\frac{\pi}{24}$

Vedclass · Answer

(B) We need to evaluate the integral $I = \int_{0}^{\frac{2}{3}} \frac{d x}{4+9 x^{2}}$.First,rewrite the integrand as $\int \frac{d x}{(2)^{2}+(3 x)^{2}}$.Let $3x = t$,then $3 dx = dt$,which implies $dx = \frac{1}{3} dt$.The indefinite integral becomes $\int \frac{1}{3} \frac{dt}{(2)^{2}+t^{2}} = \frac{1}{3} \cdot \frac{1}{2} 	an^{-1}(\frac{t}{2}) = \frac{1}{6} 	an^{-1}(\frac{3x}{2})$.Applying the limits from $0$ to $\frac{2}{3}$:$I = \left[ \frac{1}{6} 	an^{-1}(\frac{3x}{2}) ight]_{0}^{\frac{2}{3}}$$I = \frac{1}{6} 	an^{-1}(\frac{3}{2} \cdot \frac{2}{3}) - \frac{1}{6} 	an^{-1}(0)$$I = \frac{1}{6} 	an^{-1}(1) - 0$Since $	an^{-1}(1) = \frac{\pi}{4}$,we have $I = \frac{1}{6} \cdot \frac{\pi}{4} = \frac{\pi}{24}$.Thus,the correct option is $B$.

Evaluate the definite integral $\int_{0}^{\frac{2}{3}} \frac{d x}{4+9 x^{2}}$.

Similar Questions

If $\int_0^{\frac{1}{2}} \frac{x^2}{\left(1-x^2\right)^{\frac{3}{2}}} \,d x=\frac{k}{6}$, then the value of $k$ is

The value of the integral $\int \limits_{0}^{1} \frac{\sqrt{x} \, dx}{(1+x)(1+3 x)(3+x)}$ is:

$\int_0^{\pi /4} \sec x \log (\sec x + \tan x) \, dx = $

Evaluate the definite integral: $\int_{0}^{3} \sqrt{9 - x^2} dx$.

$\int_{1}^{5} (|x-3| + |1-x|) dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module