જો I_n = int_0^{frac{pi}{2}} cos^n x cos(nx) | vedclass

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

Question

(B) આપેલ છે $I_n = \int_0^{\frac{\pi}{2}} \cos^n x \cos(nx) dx$. $n=1$ માટે: $I_1 = \int_0^{\frac{\pi}{2}} \cos x \cos x dx = \int_0^{\frac{\pi}{2}} \

Vedclass · Accepted Answer

$G$.$P$.

Vedclass · Answer

(B) આપેલ છે $I_n = \int_0^{\frac{\pi}{2}} \cos^n x \cos(nx) dx$. $n=1$ માટે: $I_1 = \int_0^{\frac{\pi}{2}} \cos x \cos x dx = \int_0^{\frac{\pi}{2}} \frac{1+\cos 2x}{2} dx = \left[ \frac{x}{2} + \frac{\sin 2x}{4} \right]_0^{\frac{\pi}{2}} = \frac{\pi}{4}$. $n=2$ માટે: $I_2 = \int_0^{\frac{\pi}{2}} \cos^2 x \cos 2x dx$. $\cos 2x = 2\cos^2 x - 1$ નો ઉપયોગ કરતા, $\cos^2 x = \frac{1+\cos 2x}{2}$. $I_2 = \int_0^{\frac{\pi}{2}} \left( \frac{1+\cos 2x}{2} \right) \cos 2x dx = \frac{1}{2} \int_0^{\frac{\pi}{2}} \cos 2x dx + \frac{1}{2} \int_0^{\frac{\pi}{2}} \cos^2 2x dx$. $I_2 = \frac{1}{2} \left[ \frac{\sin 2x}{2} \right]_0^{\frac{\pi}{2}} + \frac{1}{2} \int_0^{\frac{\pi}{2}} \frac{1+\cos 4x}{2} dx = 0 + \frac{1}{4} \left[ x + \frac{\sin 4x}{4} \right]_0^{\frac{\pi}{2}} = \frac{\pi}{8}$. $n=3$ માટે: $I_3 = \int_0^{\frac{\pi}{2}} \cos^3 x \cos 3x dx$. $\cos 3x = 4\cos^3 x - 3\cos x$ નો ઉપયોગ કરતા, $\cos^3 x = \frac{\cos 3x + 3\cos x}{4}$. $I_3 = \int_0^{\frac{\pi}{2}} \left( \frac{\cos 3x + 3\cos x}{4} \right) \cos 3x dx = \frac{1}{4} \int_0^{\frac{\pi}{2}} \cos^2 3x dx + \frac{3}{4} \int_0^{\frac{\pi}{2}} \cos x \cos 3x dx$. $I_3 = \frac{1}{4} \int_0^{\frac{\pi}{2}} \frac{1+\cos 6x}{2} dx + \frac{3}{8} \int_0^{\frac{\pi}{2}} (\cos 4x + \cos 2x) dx = \frac{1}{8} [x]_0^{\frac{\pi}{2}} + 0 + 0 = \frac{\pi}{16}$. અહીં $I_1 = \frac{\pi}{4}$, $I_2 = \frac{\pi}{8}$, $I_3 = \frac{\pi}{16}$ હોવાથી, ગુણોત્તર $\frac{I_2}{I_1} = \frac{1}{2}$ અને $\frac{I_3}{I_2} = \frac{1}{2}$ છે. તેથી, $I_1, I_2, I_3, \ldots$ એ $G$.$P$. માં છે.

જો $I_n = \int_0^{\frac{\pi}{2}} \cos^n x \cos(nx) dx$ હોય, તો $I_1, I_2, I_3, \ldots$ શેમાં છે?

Similar Questions

$\int_{1 / 2}^{1 / \sqrt{2}} \frac{1}{\left(x+\sqrt{1-x^2}\right)\left(1-x^2\right)} d x=$

$\int_0^{\pi / 4} x^2 \sin 2x \, dx =$

જો $y = \sqrt{\sec x + \sqrt{\sec x + \sqrt{\sec x + \dots \infty}}} \,,$ હોય,તો $\int_{0}^{\pi/3} (2y - 1) \frac{dy}{dx} \, dx$ ની કિંમત $(\sec x > 0)$ માટે શોધો -

ધારો કે $n \in N$ માટે $a_{n} = \int_{-1}^{n} \left(1 + \frac{x}{2} + \frac{x^{2}}{3} + \ldots + \frac{x^{n-1}}{n}\right) dx$ છે. તો ગણ $\{n \in N : a_{n} \in (2, 30)\}$ ના તમામ ઘટકોનો સરવાળો $...........$ છે.

$ \int_{0}^{\frac{\pi}{2}} \frac{1}{a^{2} \sin ^{2} x+b^{2} \cos ^{2} x} d x $ ની કિંમત શોધો.

Vedclass Test Series

Exam Paper Generator

Online Exam Module