If int_{0}^{pi/2} tan^{n}(x) dx = k int_{0}^{ | vedclass

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

Question

(A) Let $I = \int_{0}^{\pi/2} 	an^{n}(x) dx$. Using the property of definite integration $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we replace $

Vedclass · Accepted Answer

$k = 1$

Vedclass · Answer

(A) Let $I = \int_{0}^{\pi/2} 	an^{n}(x) dx$. Using the property of definite integration $\int_{0}^{a} f(x) dx = \int_{0}^{a} f(a-x) dx$,we replace $x$ with $\frac{\pi}{2} - x$. $I = \int_{0}^{\pi/2} 	an^{n}(\frac{\pi}{2} - x) dx$. Since $	an(\frac{\pi}{2} - x) = \cot(x)$,we have $I = \int_{0}^{\pi/2} \cot^{n}(x) dx$. Comparing this with the given equation $\int_{0}^{\pi/2} 	an^{n}(x) dx = k \int_{0}^{\pi/2} \cot^{n}(x) dx$,we see that $I = k \cdot I$. Therefore,$k = 1$.

If $\int_{0}^{\pi/2} \tan^{n}(x) dx = k \int_{0}^{\pi/2} \cot^{n}(x) dx$,then

Similar Questions

$\int_{-\pi}^\pi \frac{2 x(1+\sin x)}{1+\cos ^2 x} d x=$

$\int_0^{\pi /2} \frac{d\theta}{1 + \tan \theta} = $

$\int_{-1}^1 \frac{\log 2 - \log(1+x)}{\sqrt{1-x^2}} dx =$

$\int_{3}^{5} \frac{\sqrt{x}}{\sqrt{8-x}+\sqrt{x}} \, dx =$

The integral $\int_{-1/2}^{1/2} \left( [x] + \log \left( \frac{1+x}{1-x} \right) \right) dx$ is equal to (where $[.]$ is the greatest integer function):

Vedclass Test Series

Exam Paper Generator

Online Exam Module