int frac{d x}{9 cos ^2 2 x+16 sin ^2 2 x}= | vedclass

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

Question

(D) Let $I = \int \frac{dx}{9 \cos^2 2x + 16 \sin^2 2x}$.Divide the numerator and denominator by $\cos^2 2x$:$I = \int \frac{\sec^2 2x dx}{9 + 16 	an

Vedclass · Accepted Answer

$\frac{1}{24} 	an ^{-1}\left(\frac{4}{3} 	an 2 xight)+c$

Vedclass · Answer

(D) Let $I = \int \frac{dx}{9 \cos^2 2x + 16 \sin^2 2x}$.Divide the numerator and denominator by $\cos^2 2x$:$I = \int \frac{\sec^2 2x dx}{9 + 16 	an^2 2x}$.Let $u = 	an 2x$, then $du = 2 \sec^2 2x dx$, which implies $\sec^2 2x dx = \frac{du}{2}$.Substituting these into the integral:$I = \int \frac{du/2}{9 + 16u^2} = \frac{1}{2} \int \frac{du}{9 + 16u^2}$.Factor out $16$ from the denominator:$I = \frac{1}{2 	imes 16} \int \frac{du}{\frac{9}{16} + u^2} = \frac{1}{32} \int \frac{du}{(\frac{3}{4})^2 + u^2}$.Using the formula $\int \frac{dx}{a^2 + x^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + C$:$I = \frac{1}{32} 	imes \frac{1}{3/4} 	an^{-1}(\frac{u}{3/4}) + C = \frac{1}{32} 	imes \frac{4}{3} 	an^{-1}(\frac{4u}{3}) + C$.$I = \frac{1}{24} 	an^{-1}(\frac{4}{3} 	an 2x) + C$.

$\int \frac{d x}{9 \cos ^2 2 x+16 \sin ^2 2 x}=$

Similar Questions

Integrate the function $\frac{1}{1+\cot x}$.

$\int(\sqrt{x+\sqrt{12x-36}}+\sqrt{x-\sqrt{12x-36}}) dx=$

The value of the integral $\int_0^1 \frac{x^b - 1}{\log x} \, dx$ is

Let $f(x) = 1 - \frac{1}{x}$, $g_2(x) = f(f(x))$, $g_3(x) = f(f(f(x)))$ and so on. If $\int x \cdot g_{2026}(x) \, dx = \int g_{2025}(x) \, dx + h(x) + c$, then $h(x) = $

$ \int \sqrt{x^{2}+2 x+5} \, dx $ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module