int x cos^2 x , dx = | vedclass

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

Question

(B) We use the identity $\cos^2 x = \frac{1 + \cos 2x}{2}$.$\int x \cos^2 x \, dx = \int x \left( \frac{1 + \cos 2x}{2} \right) \, dx = \frac{1}{2} \i

Vedclass · Accepted Answer

$\frac{x^2}{4} + \frac{1}{4}x \sin 2x + \frac{1}{8} \cos 2x + c$

Vedclass · Answer

(B) We use the identity $\cos^2 x = \frac{1 + \cos 2x}{2}$.$\int x \cos^2 x \, dx = \int x \left( \frac{1 + \cos 2x}{2} \right) \, dx = \frac{1}{2} \int x \, dx + \frac{1}{2} \int x \cos 2x \, dx$.Integrating the first part: $\frac{1}{2} \int x \, dx = \frac{x^2}{4}$.For the second part,use integration by parts: $\int u \, dv = uv - \int v \, du$,where $u = x$ and $dv = \cos 2x \, dx$. Then $du = dx$ and $v = \frac{\sin 2x}{2}$.$\frac{1}{2} \int x \cos 2x \, dx = \frac{1}{2} \left( \frac{x \sin 2x}{2} - \int \frac{\sin 2x}{2} \, dx \right) = \frac{x \sin 2x}{4} - \frac{1}{4} \int \sin 2x \, dx$.$= \frac{x \sin 2x}{4} - \frac{1}{4} \left( -\frac{\cos 2x}{2} \right) = \frac{x \sin 2x}{4} + \frac{\cos 2x}{8}$.Combining both parts: $\frac{x^2}{4} + \frac{x \sin 2x}{4} + \frac{\cos 2x}{8} + c$.

$\int x \cos^2 x \, dx = $

Similar Questions

$\int \frac{\operatorname{cosec}^2 x}{\sec ^2 x} \, dx = $ . . . . . . $+ C$.

$\int \frac{d x}{\sin ^{2} x \cos ^{2} x}$ equals

If $x > 0$ and $x \neq (2n+1) \frac{\pi}{2}$, then $\int \left(x \sqrt{x} - e^{\log(\sec x \tan x)} + \frac{3x^2 - 2x + 1}{x^2}\right) dx =$

$\int 3^{-\log _9 x^2} d x=$

$\int \frac{2 x^2 \cos \left(x^2\right)-\sin \left(x^2\right)}{x^2} d x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module