Find: int cos^{2} x , dx | vedclass

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

Question

(A) We use the trigonometric identity $\cos 2x = 2 \cos^{2} x - 1$,which implies $\cos^{2} x = \frac{1 + \cos 2x}{2}$.Substituting this into the integ

Vedclass · Accepted Answer

$\frac{x}{2} + \frac{1}{4} \sin 2x + C$

Vedclass · Answer

(A) We use the trigonometric identity $\cos 2x = 2 \cos^{2} x - 1$,which implies $\cos^{2} x = \frac{1 + \cos 2x}{2}$.Substituting this into the integral,we get:$\int \cos^{2} x \, dx = \int \frac{1 + \cos 2x}{2} \, dx$$= \frac{1}{2} \int (1 + \cos 2x) \, dx$$= \frac{1}{2} \left( \int 1 \, dx + \int \cos 2x \, dx \right)$$= \frac{1}{2} \left( x + \frac{\sin 2x}{2} \right) + C$$= \frac{x}{2} + \frac{1}{4} \sin 2x + C$

Find: $\int \cos^{2} x \, dx$

Similar Questions

$\int {\frac{{a{x^{ - 2}} + b{x^{ - 1}} + c}}{{{x^{ - 3}}}}} \,dx = $

$\int \frac{1 + \tan x \tan(x + a)}{\tan x \tan(x + a)} dx =$

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

The value of $\int \frac{dx}{x\sqrt{x^4 - 1}}$ is

$\int {\frac{{\cot x \tan x}}{{{{\sec }^2}x - 1}}} \;dx = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module