If int frac{dx}{1+3 sin^2 x} = frac{1}{2} tan | vedclass

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

Question

(A) We have the integral $I = \int \frac{dx}{1+3 \sin^2 x}$.Dividing the numerator and denominator by $\cos^2 x$,we get:$I = \int \frac{\sec^2 x dx}{\

Vedclass · Accepted Answer

$2 	an x$

Vedclass · Answer

(A) We have the integral $I = \int \frac{dx}{1+3 \sin^2 x}$.Dividing the numerator and denominator by $\cos^2 x$,we get:$I = \int \frac{\sec^2 x dx}{\sec^2 x + 3 	an^2 x} = \int \frac{\sec^2 x dx}{1 + 	an^2 x + 3 	an^2 x} = \int \frac{\sec^2 x dx}{1 + 4 	an^2 x}$.Let $t = 	an x$,then $dt = \sec^2 x dx$.The integral becomes:$I = \int \frac{dt}{1 + 4t^2} = \int \frac{dt}{1 + (2t)^2}$.Using the formula $\int \frac{du}{a^2 + u^2} = \frac{1}{a} 	an^{-1}(\frac{u}{a}) + c$,where $u = 2t$ and $du = 2dt$:$I = \frac{1}{2} \int \frac{2dt}{1 + (2t)^2} = \frac{1}{2} 	an^{-1}(2t) + c$.Substituting $t = 	an x$ back,we get:$I = \frac{1}{2} 	an^{-1}(2 	an x) + c$.Comparing this with the given expression $\frac{1}{2} 	an^{-1}(f(x)) + c$,we find $f(x) = 2 	an x$.

If $\int \frac{dx}{1+3 \sin^2 x} = \frac{1}{2} \tan^{-1}(f(x)) + c$,where $c$ is a constant of integration,then $f(x)$ is equal to

Similar Questions

If $\int {\frac{{\left( {2x + 3} \right)dx}}{{x\left( {x + 1} \right)\left( {x + 2} \right)\left( {x + 3} \right) + 1}}} = C - \frac{1}{{f(x)}}$ where $f(x)$ is of the form of $ax^2 + bx + c$ then $(a + b + c)$ equals

$\int \frac{dx}{\sqrt{e^x-1}} = 2 \tan^{-1}(f(x)) + c$,where $x > 0$ and $c$ is a constant of integration,then $f(x)$ is

Evaluate the integral: $\int \frac{x^3 \, dx}{1+x^8}$

Find the following integral: $\int \frac{dx}{\sqrt{2x-x^2}}$

$\int \frac{dx}{x^2 \sqrt{4+x^2}}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module