int_{0}^{frac{pi}{2}} frac{sin ^{1000} x}{si | vedclass

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

Question

(D) Let $ I = \int_{0}^{\frac{\pi}{2}} \frac{\sin ^{1000} x}{\sin ^{1000} x+\cos ^{1000} x} \, dx $. Using the property $ \int_{0}^{a} f(x) \, dx = \i

Vedclass · Accepted Answer

$ \frac{\pi}{4} $

Vedclass · Answer

(D) Let $ I = \int_{0}^{\frac{\pi}{2}} \frac{\sin ^{1000} x}{\sin ^{1000} x+\cos ^{1000} x} \, dx $. Using the property $ \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx $,we get: $ I = \int_{0}^{\frac{\pi}{2}} \frac{\sin ^{1000} (\frac{\pi}{2}-x)}{\sin ^{1000} (\frac{\pi}{2}-x)+\cos ^{1000} (\frac{\pi}{2}-x)} \, dx $ Since $ \sin(\frac{\pi}{2}-x) = \cos x $ and $ \cos(\frac{\pi}{2}-x) = \sin x $,we have: $ I = \int_{0}^{\frac{\pi}{2}} \frac{\cos ^{1000} x}{\cos ^{1000} x+\sin ^{1000} x} \, dx $ Adding the two expressions for $ I $: $ 2I = \int_{0}^{\frac{\pi}{2}} \frac{\sin ^{1000} x + \cos ^{1000} x}{\sin ^{1000} x+\cos ^{1000} x} \, dx $ $ 2I = \int_{0}^{\frac{\pi}{2}} 1 \, dx = [x]_{0}^{\frac{\pi}{2}} = \frac{\pi}{2} $ Therefore,$ I = \frac{\pi}{4} $.

$ \int_{0}^{\frac{\pi}{2}} \frac{\sin ^{1000} x}{\sin ^{1000} x+\cos ^{1000} x} \, dx $ is equal to

Similar Questions

If $f(x)$ is a continuous periodic function with period $T,$ then the integral $I = \int_a^{a + T} {f(x)\,dx} $ is

Let $f(x)$ be a positive function,$I_1 = \int_{-\frac{1}{2}}^1 2x f(2x(1-2x)) dx$,and $I_2 = \int_{-1}^2 f(x(1-x)) dx$. Then the value of $\frac{I_2}{I_1}$ is equal to:

$\int_0^{2\pi } {\frac{{\sin 2\theta }}{{a - b\cos \theta }}\,d\theta = } $

If $f(x) = \int\limits_0^\pi {\frac{t \sin t \, dt}{\sqrt{1 + \tan^2 x \sin^2 t}}}$ for $0 < x < \frac{\pi}{2}$,then which of the following is true?

The value of $I = \int_{-\pi / 2}^{\pi / 2} |\sin x| \, dx$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module