If int frac{dx}{(x^2+9) sqrt{x^2+16}} = frac{ | vedclass

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

Question

(A) Let $I = \int \frac{dx}{(x^2+9) \sqrt{x^2+16}}$.Put $x = 4 	an 	heta$, so $dx = 4 \sec^2 	heta \ d	heta$.Then $\sqrt{x^2+16} = \sqrt{16 	an^2

Vedclass · Accepted Answer

$\frac{\sqrt{7}}{3}$

Vedclass · Answer

(A) Let $I = \int \frac{dx}{(x^2+9) \sqrt{x^2+16}}$.Put $x = 4 	an 	heta$, so $dx = 4 \sec^2 	heta \ d	heta$.Then $\sqrt{x^2+16} = \sqrt{16 	an^2 	heta + 16} = 4 \sec 	heta$.Substituting these into the integral:$I = \int \frac{4 \sec^2 	heta \ d	heta}{(16 	an^2 	heta + 9)(4 \sec 	heta)} = \int \frac{\sec 	heta \ d	heta}{16 	an^2 	heta + 9} = \int \frac{\frac{1}{\cos 	heta} \ d	heta}{16 \frac{\sin^2 	heta}{\cos^2 	heta} + 9} = \int \frac{\cos 	heta \ d	heta}{16 \sin^2 	heta + 9 \cos^2 	heta}$.Since $\cos^2 	heta = 1 - \sin^2 	heta$, we have $16 \sin^2 	heta + 9(1 - \sin^2 	heta) = 7 \sin^2 	heta + 9$.So $I = \int \frac{\cos 	heta \ d	heta}{7 \sin^2 	heta + 9}$.Let $u = \sin 	heta$, then $du = \cos 	heta \ d	heta$.$I = \int \frac{du}{7u^2 + 9} = \frac{1}{7} \int \frac{du}{u^2 + (3/\sqrt{7})^2} = \frac{1}{7} \cdot \frac{\sqrt{7}}{3} \operatorname{Tan}^{-1} \left( \frac{u \sqrt{7}}{3} ight) + c = \frac{1}{3 \sqrt{7}} \operatorname{Tan}^{-1} \left( \frac{\sqrt{7} \sin 	heta}{3} ight) + c$.Since $x = 4 	an 	heta$, $\sin 	heta = \frac{x}{\sqrt{x^2+16}}$.Thus, $I = \frac{1}{3 \sqrt{7}} \operatorname{Tan}^{-1} \left( \frac{\sqrt{7}}{3} \cdot \frac{x}{\sqrt{x^2+16}} ight) + c$.Comparing this with the given expression, $K = \frac{\sqrt{7}}{3}$.

If $\int \frac{dx}{(x^2+9) \sqrt{x^2+16}} = \frac{1}{3 \sqrt{7}} \operatorname{Tan}^{-1} \left( K \frac{x}{\sqrt{16+x^2}} \right) + c$, then $K=$

Similar Questions

$\int \frac{1}{\sin (x-a) \sin x} \,d x=$

$\int \frac{x^2}{(x\sin x + \cos x)^2} \, dx = $

The integral $16 \int \limits_1^2 \frac{d x}{x^3(x^2+2)^2}$ is equal to

$\int \left(x^{3m} + x^{2m} + x^m\right) \left(2x^{2m} + 3x^m + 6\right)^{\frac{1}{m}} dx = $

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