int_0^{pi / 4} frac{x^2}{(x sin x+cos x)^2} d | vedclass

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

Question

(B) Let $I = \int \frac{x^2}{(x \sin x + \cos x)^2} dx$.We can rewrite the integrand as $I = \int (x \sec x) \left( \frac{x \cos x}{(x \sin x + \cos x

Vedclass · Accepted Answer

$\frac{4-\pi}{4+\pi}$

Vedclass · Answer

(B) Let $I = \int \frac{x^2}{(x \sin x + \cos x)^2} dx$.We can rewrite the integrand as $I = \int (x \sec x) \left( \frac{x \cos x}{(x \sin x + \cos x)^2} ight) dx$.Using integration by parts,let $u = x \sec x$ and $dv = \frac{x \cos x}{(x \sin x + \cos x)^2} dx$.Then $du = (\sec x + x \sec x 	an x) dx$ and $v = \frac{-1}{x \sin x + \cos x}$.$I = (x \sec x) \left( \frac{-1}{x \sin x + \cos x} ight) - \int (\sec x + x \sec x 	an x) \left( \frac{-1}{x \sin x + \cos x} ight) dx$.$I = \frac{-x \sec x}{x \sin x + \cos x} + \int \frac{\sec x (1 + x 	an x)}{x \sin x + \cos x} dx$.Since $x \sin x + \cos x = \cos x (x 	an x + 1)$,the integral becomes $\int \frac{\sec x \cdot \cos x (x 	an x + 1)}{\cos x (x 	an x + 1)} dx = \int \sec^2 x dx = 	an x$.Thus,$I = \frac{-x \sec x}{x \sin x + \cos x} + 	an x = \frac{-x + 	an x (x \sin x + \cos x)}{\cos x (x \sin x + \cos x)} = \frac{\sin x - x \cos x}{x \sin x + \cos x}$.Evaluating the definite integral: $\left[ \frac{\sin x - x \cos x}{x \sin x + \cos x} ight]_0^{\pi / 4} = \frac{\frac{1}{\sqrt{2}} - \frac{\pi}{4} \cdot \frac{1}{\sqrt{2}}}{\frac{\pi}{4} \cdot \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}} - 0 = \frac{1 - \pi/4}{1 + \pi/4} = \frac{4 - \pi}{4 + \pi}$.

$\int_0^{\pi / 4} \frac{x^2}{(x \sin x+\cos x)^2} d x=$

Similar Questions

$a > 1, \; \int_{1}^{a} [x] f'(x) dx = $

The value of $\int_0^{2 \pi} \sqrt{1+\sin \left(\frac{x}{2}\right)} d x$ is

Let $f(x) = \int \frac{\sqrt{x}}{(1+x)^2} dx$ $(x \geq 0)$. Then $f(3) - f(1)$ is equal to

If $[x]$ denotes the greatest integer less than or equal to $x,$ then the value of the integral $\int_0^2 {{x^2}[x]\,dx} $ equals (in $/3$)

If $\int_{1}^{2} \frac{dx}{(x^2 - 2x + 4)^{3/2}} = \frac{k}{k+5}$,then $k$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module