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 \s

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

$\int e^{4 x^2+8 x-4}(x+1) \cos \left(3 x^2+6 x-4\right) d x=$

If $\int \tan (x - \alpha) \cdot \tan (x + \alpha) \cdot \tan 2 x \ d x = p \log |\sec 2 x| + q \log |\sec (x + \alpha)| + r \log |\sec (x - \alpha)| + c$,then $p + q + r = . . . . . .$

Let for $f(x)=7 \tan^8 x + 7 \tan^6 x - 3 \tan^4 x - 3 \tan^2 x$,$I_1 = \int_0^{\pi/4} f(x) \, dx$ and $I_2 = \int_0^{\pi/4} x f(x) \, dx$. Then $7 I_1 + 12 I_2$ is equal to:

$\int \frac{2 \cos 2 x}{(1+\sin 2 x)(1+\cos 2 x)} d x=$

For any integer $n \geq 2$, let $I_n = \int \tan^n x \, dx$. If $I_n = \frac{1}{a} \tan^{n-1} x - b I_{n-2}$ for $n \geq 2$, then the ordered pair $(a, b)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module