int frac{1}{16-7 sin ^2 x} d x= | vedclass

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

Question

(A) To evaluate the integral $I = \int \frac{1}{16-7 \sin^2 x} dx$,divide the numerator and denominator by $\cos^2 x$:$I = \int \frac{\sec^2 x}{16 \se

Vedclass · Accepted Answer

$\frac{1}{12} 	an^{-1}\left(\frac{3 	an x}{4}ight)+c$

Vedclass · Answer

(A) To evaluate the integral $I = \int \frac{1}{16-7 \sin^2 x} dx$,divide the numerator and denominator by $\cos^2 x$:$I = \int \frac{\sec^2 x}{16 \sec^2 x - 7 	an^2 x} dx$Using $\sec^2 x = 1 + 	an^2 x$,we get:$I = \int \frac{\sec^2 x}{16(1 + 	an^2 x) - 7 	an^2 x} dx = \int \frac{\sec^2 x}{16 + 9 	an^2 x} dx$Let $	an x = t$,then $\sec^2 x dx = dt$:$I = \int \frac{dt}{16 + 9t^2} = \frac{1}{9} \int \frac{dt}{\frac{16}{9} + t^2} = \frac{1}{9} \int \frac{dt}{(\frac{4}{3})^2 + t^2}$Using the formula $\int \frac{dx}{a^2 + x^2} = \frac{1}{a} 	an^{-1}(\frac{x}{a}) + C$:$I = \frac{1}{9} 	imes \frac{1}{4/3} 	an^{-1}\left(\frac{t}{4/3}ight) + C = \frac{1}{9} 	imes \frac{3}{4} 	an^{-1}\left(\frac{3t}{4}ight) + C$$I = \frac{1}{12} 	an^{-1}\left(\frac{3 	an x}{4}ight) + C$

$\int \frac{1}{16-7 \sin ^2 x} d x=$

Similar Questions

The integral $\frac{24}{\pi} \int_{0}^{\sqrt{2}} \frac{(2-x^{2}) dx}{(2+x^{2}) \sqrt{4+x^{4}}}$ is equal to

If $\int \frac{5 \tan x}{\tan x-2} d x=x+a \log |\sin x-2 \cos x|+c$,then find the value of $a$ (where $c$ is the constant of integration).

$\int \frac{1}{(x+2) \sqrt{x^2+x+2}} \, dx =$

Observe the following statements :
$A: \int \left(\frac{x^2-1}{x^2}\right) e^{\frac{x^2+1}{x}} d x = e^{\frac{x^2+1}{x}} + c$
$R: \int f^{\prime}(x) e^{f(x)} d x = f(x) + c$
Then which of the following is true?

$\int \frac{\cos 7x - \cos 8x}{1 + 2 \cos 5x} dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module